{"article_info":{"article_id":110990,"cat_id":941,"cat_title":"Ôn tập hè Toán lớp 6","publish_time":1655888520,"cat_id2":941,"is_link":0,"thumbnail":"","title":"Dạng 6. Dãy phân số viết theo quy luật Chủ đề 6 Ôn hè Toán 6","label":"Dạng 6. Dãy phân số viết theo quy luật ","alias":"dang-6-day-phan-so-viet-theo-quy-luat-chu-de-6-on-he-toan-6","introtext":"Phát hiện quy luật của dãy số","content":"<p><strong class=\"sub_question_head\" id=\"sub_question_head_1\">Lý thuyết</strong></p><p>Phát hiện quy luật của dãy số</p>\r\n<p>Dạng tổng quát: \\(\\dfrac{k}{{\\left( {n - k} \\right).n}} = \\dfrac{{n - \\left( {n - k} \\right)}}{{\\left( {n - k} \\right).n}} = \\dfrac{n}{{\\left( {n - k} \\right).n}} - \\dfrac{{n - k}}{{\\left( {n - k} \\right).n}} = \\dfrac{1}{{n - k}} - \\dfrac{1}{n}\\)</p>\r\n<p>Áp dụng phương pháp khử liên tiếp: Viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.<strong></strong></p><hr/><p><strong class=\"sub_question_head\" id=\"sub_question_head_2\">Bài tập</strong></p><p><strong>Bài 1:</strong></p>\r\n<p>Tính:</p>\r\n<p>a) A = \\(2017:\\left( {\\dfrac{1}{{1.2}} + \\dfrac{1}{{2.3}} + \\dfrac{1}{{3.4}} + ... + \\dfrac{1}{{2017.2018}}} \\right)\\)</p>\r\n<p>b) \\(B = \\dfrac{3}{{2.5}} + \\dfrac{3}{{5.8}} + \\dfrac{3}{{8.11}} +  \\ldots  + \\dfrac{3}{{2016.2019}}\\)</p>\r\n<p>c) \\(C = \\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}\\)                                 </p>\r\n<p>d) \\(D = \\dfrac{7}{{1.9}} + \\dfrac{7}{{9.17}} + \\dfrac{7}{{17.25}} +  \\ldots  + \\dfrac{7}{{2011.2019}}\\)</p>\r\n<p>e) \\(E = \\dfrac{{{3^2}}}{{1.4}} + \\dfrac{{{3^2}}}{{4.7}} + \\dfrac{{{3^2}}}{{7.10}} +  \\ldots  + \\dfrac{{{3^2}}}{{2017.2020}}\\)<strong>                            </strong></p>\r\n<p>f) \\(F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tính các tổng sau:</p>\r\n<p>a) \\(A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)<strong>                                                                                       </strong></p>\r\n<p>b) \\(B = 1 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{1}{{16}} + \\dfrac{1}{{32}} +  \\ldots  + \\dfrac{1}{{2048}}\\)<strong></strong></p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>a) Tính tổng sau: \\(A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\)</p>\r\n<p>b) Chứng minh rằng biểu thức \\(B\\) có giá trị bằng \\(\\dfrac{1}{2}\\) với \\(B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}.\\)</p>\r\n<p> </p>\r\n<p><strong>Hướng dẫn giải chi tiết</strong></p>\r\n<p><strong>Bài 1:</strong></p>\r\n<p>Tính:</p>\r\n<p>a) A = \\(2017:\\left( {\\dfrac{1}{{1.2}} + \\dfrac{1}{{2.3}} + \\dfrac{1}{{3.4}} + ... + \\dfrac{1}{{2017.2018}}} \\right)\\)</p>\r\n<p>b) b) \\(B = \\dfrac{3}{{2.5}} + \\dfrac{3}{{5.8}} + \\dfrac{3}{{8.11}} +  \\ldots  + \\dfrac{3}{{2016.2019}}\\)</p>\r\n<p>c) \\(C = \\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}\\)                                 </p>\r\n<p>d) \\(D = \\dfrac{7}{{1.9}} + \\dfrac{7}{{9.17}} + \\dfrac{7}{{17.25}} +  \\ldots  + \\dfrac{7}{{2011.2019}}\\)</p>\r\n<p>e) \\(E = \\dfrac{{{3^2}}}{{1.4}} + \\dfrac{{{3^2}}}{{4.7}} + \\dfrac{{{3^2}}}{{7.10}} +  \\ldots  + \\dfrac{{{3^2}}}{{2017.2020}}\\)<strong>                            </strong></p>\r\n<p>f) \\(F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Nhận xét: Tử số bằng hiệu của các thừa số ở mẫu.</p>\r\n<p>Dạng tổng quát: \\(\\dfrac{k}{{\\left( {n - k} \\right).n}} = \\dfrac{{n - \\left( {n - k} \\right)}}{{\\left( {n - k} \\right).n}} = \\dfrac{n}{{\\left( {n - k} \\right).n}} - \\dfrac{{n - k}}{{\\left( {n - k} \\right).n}} = \\dfrac{1}{{n - k}} - \\dfrac{1}{n}\\)</p>\r\n<p>Áp dụng phương pháp khử liên tiếp: Viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.<strong></strong></p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>\\(2017:\\left( {\\dfrac{1}{{1.2}} + \\dfrac{1}{{2.3}} + \\dfrac{1}{{3.4}} + ... + \\dfrac{1}{{2017.2018}}} \\right)\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{ = 2017:\\left( {1 - \\dfrac{1}{2} + \\dfrac{1}{2} - \\dfrac{1}{3} + ... + \\dfrac{1}{{2017}} - \\dfrac{1}{{2018}}} \\right)}\\\\{ = 2017:\\left( {1 - \\dfrac{1}{{2018}}} \\right)}\\\\{ = 2017:\\dfrac{{2017}}{{2018}}}\\\\{ = 2017.\\dfrac{{2018}}{{2017}} = 2018.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 2}}{3}\\)</p>\r\n<p>b) \\(B = \\dfrac{3}{{2.5}} + \\dfrac{3}{{5.8}} + \\dfrac{3}{{8.11}} +  \\ldots  + \\dfrac{3}{{2016.2019}}\\)</p>\r\n<p>        \\(\\begin{array}{l} = \\dfrac{{5 - 2}}{{2.5}} + \\dfrac{{8 - 5}}{{5.8}} + \\dfrac{{11 - 8}}{{8.11}} +  \\ldots  + \\dfrac{{2019 - 2016}}{{2016.2019}}\\\\\\, = \\dfrac{5}{{2.5}} - \\dfrac{2}{{2.5}} + \\dfrac{8}{{5.8}} - \\dfrac{5}{{5.8}} + \\dfrac{{11}}{{8.11}} - \\dfrac{8}{{8.11}} +  \\ldots  + \\dfrac{{2019}}{{2016.2019}} - \\dfrac{{2016}}{{2016.2019}}\\\\\\, = \\dfrac{1}{2} - \\dfrac{1}{5} + \\dfrac{1}{5} - \\dfrac{1}{8} + \\dfrac{1}{8} - \\dfrac{1}{{11}} +  \\ldots  + \\dfrac{1}{{2016}} - \\dfrac{1}{{2019}}\\\\\\, = \\dfrac{1}{2} - \\dfrac{1}{{2019}}\\\\\\, = \\dfrac{{2019 - 2}}{{2.2019}}\\\\\\, = \\dfrac{{2017}}{{4038}}.\\end{array}\\)</p>\r\n<p>c) \\(C = \\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}\\)</p>\r\n<p>Xét từng phân số ta thấy: Hiệu 2 thừa số ở mẫu bằng \\(6\\) \\( \\Rightarrow \\) Nhân cả 2 vế của biểu thức với \\(3\\).</p>\r\n<p>\\(\\begin{array}{l} \\Rightarrow 3C = 3 \\cdot \\left( {\\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{6}{{1.7}} + \\dfrac{6}{{7.13}} + \\dfrac{6}{{13.19}} +  \\ldots  + \\dfrac{6}{{2013.2019}}\\end{array}\\)</p>\r\n<p>\\(\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\left( {\\dfrac{1}{1} - \\dfrac{1}{7}} \\right) + \\left( {\\dfrac{1}{7} - \\dfrac{1}{{13}}} \\right) + \\left( {\\dfrac{1}{{13}} - \\dfrac{1}{{19}}} \\right) +  \\ldots  + \\left( {\\dfrac{1}{{2013}} - \\dfrac{1}{{2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{1}{1} - \\dfrac{1}{7} + \\dfrac{1}{7} - \\dfrac{1}{{13}} + \\dfrac{1}{{13}} - \\dfrac{1}{{19}} +  \\ldots  + \\dfrac{1}{{2013}} - \\dfrac{1}{{2019}}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = 1 - \\dfrac{1}{{2019}} = \\dfrac{{2018}}{{2019}}\\end{array}\\)</p>\r\n<p>\\( \\Rightarrow 3C = \\dfrac{{2018}}{{2019}} \\Rightarrow C = \\dfrac{{2018}}{{2019}}:3 = \\dfrac{{2018}}{{2019}} \\cdot \\dfrac{1}{3} = \\dfrac{{2018}}{{6057}}\\)</p>\r\n<p>d) \\(D = \\dfrac{7}{{1.9}} + \\dfrac{7}{{9.17}} + \\dfrac{7}{{17.25}} +  \\ldots  + \\dfrac{7}{{2011.2019}}\\)</p>\r\n<p>\\(\\begin{array}{l} \\Rightarrow D = 7 \\cdot \\dfrac{8}{8} \\cdot \\left( {\\dfrac{1}{{1.9}} + \\dfrac{1}{{9.17}} + \\dfrac{1}{{17.25}} +  \\ldots  + \\dfrac{1}{{2011.2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\left( {\\dfrac{8}{{1.9}} + \\dfrac{8}{{9.17}} + \\dfrac{8}{{17.25}} +  \\ldots  + \\dfrac{8}{{2011.2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\left( {1 - \\dfrac{1}{9} + \\dfrac{1}{9} - \\dfrac{1}{{17}} + \\dfrac{1}{{17}} - \\dfrac{1}{{25}} +  \\ldots  + \\dfrac{1}{{2011}} - \\dfrac{1}{{2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\left( {1 - \\dfrac{1}{{2019}}} \\right) = \\dfrac{7}{8} \\cdot \\left( {\\dfrac{{2019}}{{2019}} - \\dfrac{1}{{2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\dfrac{{2018}}{{2019}} = \\dfrac{{7.1009}}{{4.2019}} = \\dfrac{{7063}}{{8076}}\\end{array}\\)</p>\r\n<p>Vậy \\(D = \\dfrac{{7063}}{{8076}}.\\)</p>\r\n<p>e) \\(E = \\dfrac{{{3^2}}}{{1.4}} + \\dfrac{{{3^2}}}{{4.7}} + \\dfrac{{{3^2}}}{{7.10}} +  \\ldots  + \\dfrac{{{3^2}}}{{2017.2020}}\\)</p>\r\n<p>       \\(\\begin{array}{l} = \\dfrac{{3.3}}{{1.4}} + \\dfrac{{3.3}}{{4.7}} + \\dfrac{{3.3}}{{7.10}} +  \\ldots  + \\dfrac{{3.3}}{{2017.2020}}\\\\ = 3 \\cdot \\left( {\\dfrac{3}{{1.4}} + \\dfrac{3}{{4.7}} + \\dfrac{3}{{7.10}} +  \\ldots  + \\dfrac{3}{{2017.2020}}} \\right)\\\\ = 3 \\cdot \\left( {\\dfrac{1}{1} - \\dfrac{1}{4} + \\dfrac{1}{4} - \\dfrac{1}{7} + \\dfrac{1}{7} - \\dfrac{1}{{10}} +  \\ldots  + \\dfrac{1}{{2017}} - \\dfrac{1}{{2020}}} \\right)\\\\ = 3 \\cdot \\left( {1 - \\dfrac{1}{{2020}}} \\right) = 3 \\cdot \\left( {\\dfrac{{2020}}{{2020}} - \\dfrac{1}{{2020}}} \\right)\\\\ = 3 \\cdot \\dfrac{{2019}}{{2020}} = \\dfrac{{6057}}{{2020}}\\end{array}\\)</p>\r\n<p>Vậy \\(E = \\dfrac{{6057}}{{2020}} \\cdot \\)</p>\r\n<p>f) \\(F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p>Ta xét:</p>\r\n<p>\\(\\dfrac{2}{{1.2.3}} = \\dfrac{{3 - 1}}{{1.2.3}} = \\dfrac{3}{{1.2.3}} - \\dfrac{1}{{1.2.3}} = \\dfrac{1}{{1.2}} - \\dfrac{1}{{2.3}}\\)</p>\r\n<p>\\(\\dfrac{2}{{2.3.4}} = \\dfrac{{4 - 2}}{{2.3.4}} = \\dfrac{4}{{2.3.4}} - \\dfrac{2}{{2.3.4}} = \\dfrac{1}{{2.3}} - \\dfrac{1}{{3.4}}\\)</p>\r\n<p>\\(........\\)</p>\r\n<p>\\(\\dfrac{2}{{18.19.20}} = \\dfrac{{20 - 18}}{{18.19.20}}\\)\\( = \\dfrac{{20}}{{18.19.20}} - \\dfrac{{18}}{{18.19.20}}\\)\\( = \\dfrac{1}{{18.19}} - \\dfrac{1}{{19.20}}\\)</p>\r\n<p>Tổng quát: \\(\\dfrac{1}{{n.\\left( {n + 1} \\right)}} - \\dfrac{1}{{\\left( {n + 1} \\right)\\left( {n + 2} \\right)}} = \\dfrac{2}{{n\\left( {n + 1} \\right)\\left( {n + 2} \\right)}}\\)</p>\r\n<p>\\( \\Rightarrow F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p>\\( \\Rightarrow 2F = \\dfrac{2}{{1.2.3}} + \\dfrac{2}{{2.3.4}} + \\dfrac{2}{{3.4.5}} +  \\ldots  + \\dfrac{2}{{18.19.20}}\\)</p>\r\n<p>     \\(\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{1}{{1.2}} - \\dfrac{1}{{2.3}} + \\dfrac{1}{{2.3}} - \\dfrac{1}{{3.4}} + \\dfrac{1}{{3.4}} - \\dfrac{1}{{4.5}} +  \\ldots  + \\dfrac{1}{{18.19}} - \\dfrac{1}{{19.20}}\\)</p>\r\n<p>     \\(\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{1}{{1.2}} - \\dfrac{1}{{19.20}} = \\dfrac{{19.10 - 1}}{{19.20}} = \\dfrac{{190 - 1}}{{380}} = \\dfrac{{189}}{{380}}\\)</p>\r\n<p>\\( \\Rightarrow F = \\dfrac{{189}}{{380}}:2 = \\dfrac{{189}}{{380}} \\cdot \\dfrac{1}{2} = \\dfrac{{189}}{{760}}\\)</p>\r\n<p>Vậy \\(F = \\dfrac{{189}}{{760}} \\cdot \\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tính các tổng sau:</p>\r\n<p>a) \\(A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)<strong>                                                                                       </strong></p>\r\n<p>b) \\(B = 1 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{1}{{16}} + \\dfrac{1}{{32}} +  \\ldots  + \\dfrac{1}{{2048}}\\)<strong></strong></p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Xét các phân số có tử bằng nhau và có mẫu là lũy thừa tăng dần của cùng 1 cơ số thì ta nhân cả 2 vế với đúng cơ số đó. Trường hợp tổng quát:</p>\r\n<p>\\(A = \\dfrac{1}{{{a^1}}} + \\dfrac{1}{{{a^2}}} + \\dfrac{1}{{{a^3}}} +  \\ldots  + \\dfrac{1}{{{a^n}}}\\)\\( \\Rightarrow A.a = a\\left( {\\dfrac{1}{{{a^1}}} + \\dfrac{1}{{{a^2}}} + \\dfrac{1}{{{a^3}}} +  \\ldots  + \\dfrac{1}{{{a^n}}}} \\right)\\)\\( = 1 + \\dfrac{1}{a} +  \\ldots  + \\dfrac{1}{{{a^{n - 1}}}}\\)</p>\r\n<p> </p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>a) \\(A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2A = 2 \\cdot \\left( {\\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}} \\right)\\)</p>\r\n<p>\\( \\Rightarrow 2A = 2 \\cdot \\dfrac{1}{2} + 2 \\cdot \\dfrac{1}{{{2^2}}} + 2 \\cdot \\dfrac{1}{{{2^3}}} + 2 \\cdot \\dfrac{1}{{{2^4}}} +  \\ldots  + 2 \\cdot \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2A = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{2019}}}}\\)</p>\r\n<p>\\(\\,\\,\\,\\,\\,\\,\\,\\,\\,A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2A - A = \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{2019}}}}} \\right) - \\left( {\\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}} \\right)\\)</p>\r\n<p>\\( \\Rightarrow A = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{2019}}}} - \\dfrac{1}{2} - \\dfrac{1}{{{2^2}}} - \\dfrac{1}{{{2^3}}} - \\dfrac{1}{{{2^4}}} -  \\ldots  - \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow A = 1 - \\dfrac{1}{{{2^{2020}}}} = \\dfrac{{{2^{2020}} - 1}}{{{2^{2020}}}}\\)</p>\r\n<p>Vậy \\(A = \\dfrac{{{2^{2020}} - 1}}{{{2^{2020}}}}\\).</p>\r\n<p>b) \\(B = 1 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{1}{{16}} +  \\ldots  + \\dfrac{1}{{2048}}\\) \\( = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{11}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B = 2 \\cdot \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{11}}}}} \\right)\\)</p>\r\n<p>\\( \\Rightarrow 2B = 2.1 + 2 \\cdot \\dfrac{1}{2} + 2 \\cdot \\dfrac{1}{{{2^2}}} + 2 \\cdot \\dfrac{1}{{{2^3}}} + 2 \\cdot \\dfrac{1}{{{2^4}}} +  \\ldots  + 2 \\cdot \\dfrac{1}{{{2^{11}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B = 2 + 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{10}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B = 3 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{10}}}}\\)</p>\r\n<p>\\(\\,\\,\\,\\,\\,\\,\\,\\,\\,B = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{11}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B - B = 2 - \\dfrac{1}{{{2^{11}}}}\\)\\( \\Rightarrow B = \\dfrac{{{2^{12}} - 1}}{{{2^{11}}}}\\);</p>\r\n<p>Vậy \\(B = \\dfrac{{{2^{12}} - 1}}{{{2^{11}}}} \\cdot \\)</p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>a) Tính tổng sau: \\(A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\)</p>\r\n<p>b) Chứng minh rằng biểu thức \\(B\\) có giá trị bằng \\(\\dfrac{1}{2}\\) với \\(B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}.\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>+) Áp dụng quy tắc dấu ngoặc, nhóm các hạng tử.</p>\r\n<p>+) Áp dụng công thức tính tổng của 1 dãy các số tự nhiên liên tiếp: \\(1 + 2 +  \\ldots  + n = \\dfrac{{n + 1}}{2} \\cdot n = \\dfrac{{n.\\left( {n + 1} \\right)}}{2}\\)</p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>a) \\(A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\)</p>\r\n<p>Ta có:</p>\r\n<p>\\(\\begin{array}{l}A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\\\\\,\\,\\,\\,\\, = \\dfrac{{1 + 1 + 2 + 1 + 2 + 3 +  \\ldots  + 1 + 2 + 3 +  \\ldots 2020}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\\\\\,\\,\\,\\,\\, = \\dfrac{{\\left( {1 + 1 + 1 +  \\ldots  + 1} \\right) + \\left( {2 + 2 +  \\ldots 2} \\right) + \\left( {3 + 3 + 3 + 3 \\ldots } \\right) +  \\ldots  + 2020}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\end{array}\\)<strong></strong></p>\r\n<p>\\(\\begin{array}{l}\\,\\,\\,\\,\\, = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\\\\\,\\,\\,\\,\\, = 1\\end{array}\\)<strong></strong></p>\r\n<p>b) Chứng minh rằng biểu thức \\(B\\) có giá trị bằng \\(\\dfrac{1}{2}\\) với \\(B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}.\\)</p>\r\n<p>Với biểu thức \\(B\\), xét tử số ta có:</p>\r\n<p>\\(\\,\\,\\,\\,1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1\\)</p>\r\n<p>\\( = 1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)\\)</p>\r\n<p>\\( = \\dfrac{{0 + 1}}{2} \\cdot 2 + \\dfrac{{1 + 2}}{2} \\cdot 2 + \\dfrac{{1 + 3}}{2} \\cdot 3 +  \\ldots  + \\dfrac{{1 + 2020}}{2} \\cdot 2020\\)</p>\r\n<p>\\( = \\dfrac{1}{2} \\cdot 2 + \\dfrac{3}{2} \\cdot 2 + \\dfrac{4}{2} \\cdot 3 +  \\ldots  + \\dfrac{{2021}}{2} \\cdot 2020\\)</p>\r\n<p>\\( = \\dfrac{{1.2}}{2} + \\dfrac{{2.3}}{2} + \\dfrac{{3.4}}{2} +  \\ldots  + \\dfrac{{2020.2021}}{2}\\)</p>\r\n<p>\\( = \\dfrac{1}{2} \\cdot \\left( {1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021} \\right)\\)</p>\r\n<p>\\(\\begin{array}{l} \\Rightarrow B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{{\\dfrac{1}{2} \\cdot \\left( {1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021} \\right)}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}} = \\dfrac{1}{2}.\\end{array}\\)</p>\r\n<p>Vậy \\(B = \\dfrac{1}{2} \\cdot \\)</p>","common_article":"","display_type":2,"article_type":1,"cmt_count":0,"zip_status":1,"last_modified_time":1655888742,"last_zip_time":1655888742,"zip_link":"https://offline.loigiaihay.com/articles/941/110990/110990_1655888742.zip","picture":"","youtube_id":"","url_upload_pdf":"https://img.loigiaihay.com/picture/share_img/2022/0622/dang-6-day-phan-so-viet-theo-quy-luat-chu-de-6-on-he-toan-6-1655888793.pdf","video_start_time":"0","quizlet_id":"0","ref_article_id":0,"readmore_display_type":0,"wiki":0,"sub_articles":[{"id":"440035","group_title":"","sub_article_link":"0","order_in_article":"1","title":"Lý thuyết","content":"<p>Phát hiện quy luật của dãy số</p>\r\n<p>Dạng tổng quát: \\(\\dfrac{k}{{\\left( {n - k} \\right).n}} = \\dfrac{{n - \\left( {n - k} \\right)}}{{\\left( {n - k} \\right).n}} = \\dfrac{n}{{\\left( {n - k} \\right).n}} - \\dfrac{{n - k}}{{\\left( {n - k} \\right).n}} = \\dfrac{1}{{n - k}} - \\dfrac{1}{n}\\)</p>\r\n<p>Áp dụng phương pháp khử liên tiếp: Viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.<strong></strong></p>","youtube_id":"","video_start_time":"0"},{"id":"440036","group_title":"","sub_article_link":"0","order_in_article":"2","title":"Bài tập","content":"<p><strong>Bài 1:</strong></p>\r\n<p>Tính:</p>\r\n<p>a) A = \\(2017:\\left( {\\dfrac{1}{{1.2}} + \\dfrac{1}{{2.3}} + \\dfrac{1}{{3.4}} + ... + \\dfrac{1}{{2017.2018}}} \\right)\\)</p>\r\n<p>b) \\(B = \\dfrac{3}{{2.5}} + \\dfrac{3}{{5.8}} + \\dfrac{3}{{8.11}} +  \\ldots  + \\dfrac{3}{{2016.2019}}\\)</p>\r\n<p>c) \\(C = \\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}\\)                                 </p>\r\n<p>d) \\(D = \\dfrac{7}{{1.9}} + \\dfrac{7}{{9.17}} + \\dfrac{7}{{17.25}} +  \\ldots  + \\dfrac{7}{{2011.2019}}\\)</p>\r\n<p>e) \\(E = \\dfrac{{{3^2}}}{{1.4}} + \\dfrac{{{3^2}}}{{4.7}} + \\dfrac{{{3^2}}}{{7.10}} +  \\ldots  + \\dfrac{{{3^2}}}{{2017.2020}}\\)<strong>                            </strong></p>\r\n<p>f) \\(F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tính các tổng sau:</p>\r\n<p>a) \\(A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)<strong>                                                                                       </strong></p>\r\n<p>b) \\(B = 1 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{1}{{16}} + \\dfrac{1}{{32}} +  \\ldots  + \\dfrac{1}{{2048}}\\)<strong></strong></p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>a) Tính tổng sau: \\(A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\)</p>\r\n<p>b) Chứng minh rằng biểu thức \\(B\\) có giá trị bằng \\(\\dfrac{1}{2}\\) với \\(B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}.\\)</p>\r\n<p> </p>\r\n<p><strong>Hướng dẫn giải chi tiết</strong></p>\r\n<p><strong>Bài 1:</strong></p>\r\n<p>Tính:</p>\r\n<p>a) A = \\(2017:\\left( {\\dfrac{1}{{1.2}} + \\dfrac{1}{{2.3}} + \\dfrac{1}{{3.4}} + ... + \\dfrac{1}{{2017.2018}}} \\right)\\)</p>\r\n<p>b) b) \\(B = \\dfrac{3}{{2.5}} + \\dfrac{3}{{5.8}} + \\dfrac{3}{{8.11}} +  \\ldots  + \\dfrac{3}{{2016.2019}}\\)</p>\r\n<p>c) \\(C = \\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}\\)                                 </p>\r\n<p>d) \\(D = \\dfrac{7}{{1.9}} + \\dfrac{7}{{9.17}} + \\dfrac{7}{{17.25}} +  \\ldots  + \\dfrac{7}{{2011.2019}}\\)</p>\r\n<p>e) \\(E = \\dfrac{{{3^2}}}{{1.4}} + \\dfrac{{{3^2}}}{{4.7}} + \\dfrac{{{3^2}}}{{7.10}} +  \\ldots  + \\dfrac{{{3^2}}}{{2017.2020}}\\)<strong>                            </strong></p>\r\n<p>f) \\(F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Nhận xét: Tử số bằng hiệu của các thừa số ở mẫu.</p>\r\n<p>Dạng tổng quát: \\(\\dfrac{k}{{\\left( {n - k} \\right).n}} = \\dfrac{{n - \\left( {n - k} \\right)}}{{\\left( {n - k} \\right).n}} = \\dfrac{n}{{\\left( {n - k} \\right).n}} - \\dfrac{{n - k}}{{\\left( {n - k} \\right).n}} = \\dfrac{1}{{n - k}} - \\dfrac{1}{n}\\)</p>\r\n<p>Áp dụng phương pháp khử liên tiếp: Viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.<strong></strong></p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>\\(2017:\\left( {\\dfrac{1}{{1.2}} + \\dfrac{1}{{2.3}} + \\dfrac{1}{{3.4}} + ... + \\dfrac{1}{{2017.2018}}} \\right)\\)</p>\r\n<p>\\(\\begin{array}{*{20}{l}}{ = 2017:\\left( {1 - \\dfrac{1}{2} + \\dfrac{1}{2} - \\dfrac{1}{3} + ... + \\dfrac{1}{{2017}} - \\dfrac{1}{{2018}}} \\right)}\\\\{ = 2017:\\left( {1 - \\dfrac{1}{{2018}}} \\right)}\\\\{ = 2017:\\dfrac{{2017}}{{2018}}}\\\\{ = 2017.\\dfrac{{2018}}{{2017}} = 2018.}\\end{array}\\)</p>\r\n<p>Vậy \\(x = \\dfrac{{ - 2}}{3}\\)</p>\r\n<p>b) \\(B = \\dfrac{3}{{2.5}} + \\dfrac{3}{{5.8}} + \\dfrac{3}{{8.11}} +  \\ldots  + \\dfrac{3}{{2016.2019}}\\)</p>\r\n<p>        \\(\\begin{array}{l} = \\dfrac{{5 - 2}}{{2.5}} + \\dfrac{{8 - 5}}{{5.8}} + \\dfrac{{11 - 8}}{{8.11}} +  \\ldots  + \\dfrac{{2019 - 2016}}{{2016.2019}}\\\\\\, = \\dfrac{5}{{2.5}} - \\dfrac{2}{{2.5}} + \\dfrac{8}{{5.8}} - \\dfrac{5}{{5.8}} + \\dfrac{{11}}{{8.11}} - \\dfrac{8}{{8.11}} +  \\ldots  + \\dfrac{{2019}}{{2016.2019}} - \\dfrac{{2016}}{{2016.2019}}\\\\\\, = \\dfrac{1}{2} - \\dfrac{1}{5} + \\dfrac{1}{5} - \\dfrac{1}{8} + \\dfrac{1}{8} - \\dfrac{1}{{11}} +  \\ldots  + \\dfrac{1}{{2016}} - \\dfrac{1}{{2019}}\\\\\\, = \\dfrac{1}{2} - \\dfrac{1}{{2019}}\\\\\\, = \\dfrac{{2019 - 2}}{{2.2019}}\\\\\\, = \\dfrac{{2017}}{{4038}}.\\end{array}\\)</p>\r\n<p>c) \\(C = \\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}\\)</p>\r\n<p>Xét từng phân số ta thấy: Hiệu 2 thừa số ở mẫu bằng \\(6\\) \\( \\Rightarrow \\) Nhân cả 2 vế của biểu thức với \\(3\\).</p>\r\n<p>\\(\\begin{array}{l} \\Rightarrow 3C = 3 \\cdot \\left( {\\dfrac{2}{{1.7}} + \\dfrac{2}{{7.13}} + \\dfrac{2}{{13.19}} +  \\ldots  + \\dfrac{2}{{2013.2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{6}{{1.7}} + \\dfrac{6}{{7.13}} + \\dfrac{6}{{13.19}} +  \\ldots  + \\dfrac{6}{{2013.2019}}\\end{array}\\)</p>\r\n<p>\\(\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\left( {\\dfrac{1}{1} - \\dfrac{1}{7}} \\right) + \\left( {\\dfrac{1}{7} - \\dfrac{1}{{13}}} \\right) + \\left( {\\dfrac{1}{{13}} - \\dfrac{1}{{19}}} \\right) +  \\ldots  + \\left( {\\dfrac{1}{{2013}} - \\dfrac{1}{{2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{1}{1} - \\dfrac{1}{7} + \\dfrac{1}{7} - \\dfrac{1}{{13}} + \\dfrac{1}{{13}} - \\dfrac{1}{{19}} +  \\ldots  + \\dfrac{1}{{2013}} - \\dfrac{1}{{2019}}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = 1 - \\dfrac{1}{{2019}} = \\dfrac{{2018}}{{2019}}\\end{array}\\)</p>\r\n<p>\\( \\Rightarrow 3C = \\dfrac{{2018}}{{2019}} \\Rightarrow C = \\dfrac{{2018}}{{2019}}:3 = \\dfrac{{2018}}{{2019}} \\cdot \\dfrac{1}{3} = \\dfrac{{2018}}{{6057}}\\)</p>\r\n<p>d) \\(D = \\dfrac{7}{{1.9}} + \\dfrac{7}{{9.17}} + \\dfrac{7}{{17.25}} +  \\ldots  + \\dfrac{7}{{2011.2019}}\\)</p>\r\n<p>\\(\\begin{array}{l} \\Rightarrow D = 7 \\cdot \\dfrac{8}{8} \\cdot \\left( {\\dfrac{1}{{1.9}} + \\dfrac{1}{{9.17}} + \\dfrac{1}{{17.25}} +  \\ldots  + \\dfrac{1}{{2011.2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\left( {\\dfrac{8}{{1.9}} + \\dfrac{8}{{9.17}} + \\dfrac{8}{{17.25}} +  \\ldots  + \\dfrac{8}{{2011.2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\left( {1 - \\dfrac{1}{9} + \\dfrac{1}{9} - \\dfrac{1}{{17}} + \\dfrac{1}{{17}} - \\dfrac{1}{{25}} +  \\ldots  + \\dfrac{1}{{2011}} - \\dfrac{1}{{2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\left( {1 - \\dfrac{1}{{2019}}} \\right) = \\dfrac{7}{8} \\cdot \\left( {\\dfrac{{2019}}{{2019}} - \\dfrac{1}{{2019}}} \\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{7}{8} \\cdot \\dfrac{{2018}}{{2019}} = \\dfrac{{7.1009}}{{4.2019}} = \\dfrac{{7063}}{{8076}}\\end{array}\\)</p>\r\n<p>Vậy \\(D = \\dfrac{{7063}}{{8076}}.\\)</p>\r\n<p>e) \\(E = \\dfrac{{{3^2}}}{{1.4}} + \\dfrac{{{3^2}}}{{4.7}} + \\dfrac{{{3^2}}}{{7.10}} +  \\ldots  + \\dfrac{{{3^2}}}{{2017.2020}}\\)</p>\r\n<p>       \\(\\begin{array}{l} = \\dfrac{{3.3}}{{1.4}} + \\dfrac{{3.3}}{{4.7}} + \\dfrac{{3.3}}{{7.10}} +  \\ldots  + \\dfrac{{3.3}}{{2017.2020}}\\\\ = 3 \\cdot \\left( {\\dfrac{3}{{1.4}} + \\dfrac{3}{{4.7}} + \\dfrac{3}{{7.10}} +  \\ldots  + \\dfrac{3}{{2017.2020}}} \\right)\\\\ = 3 \\cdot \\left( {\\dfrac{1}{1} - \\dfrac{1}{4} + \\dfrac{1}{4} - \\dfrac{1}{7} + \\dfrac{1}{7} - \\dfrac{1}{{10}} +  \\ldots  + \\dfrac{1}{{2017}} - \\dfrac{1}{{2020}}} \\right)\\\\ = 3 \\cdot \\left( {1 - \\dfrac{1}{{2020}}} \\right) = 3 \\cdot \\left( {\\dfrac{{2020}}{{2020}} - \\dfrac{1}{{2020}}} \\right)\\\\ = 3 \\cdot \\dfrac{{2019}}{{2020}} = \\dfrac{{6057}}{{2020}}\\end{array}\\)</p>\r\n<p>Vậy \\(E = \\dfrac{{6057}}{{2020}} \\cdot \\)</p>\r\n<p>f) \\(F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p>Ta xét:</p>\r\n<p>\\(\\dfrac{2}{{1.2.3}} = \\dfrac{{3 - 1}}{{1.2.3}} = \\dfrac{3}{{1.2.3}} - \\dfrac{1}{{1.2.3}} = \\dfrac{1}{{1.2}} - \\dfrac{1}{{2.3}}\\)</p>\r\n<p>\\(\\dfrac{2}{{2.3.4}} = \\dfrac{{4 - 2}}{{2.3.4}} = \\dfrac{4}{{2.3.4}} - \\dfrac{2}{{2.3.4}} = \\dfrac{1}{{2.3}} - \\dfrac{1}{{3.4}}\\)</p>\r\n<p>\\(........\\)</p>\r\n<p>\\(\\dfrac{2}{{18.19.20}} = \\dfrac{{20 - 18}}{{18.19.20}}\\)\\( = \\dfrac{{20}}{{18.19.20}} - \\dfrac{{18}}{{18.19.20}}\\)\\( = \\dfrac{1}{{18.19}} - \\dfrac{1}{{19.20}}\\)</p>\r\n<p>Tổng quát: \\(\\dfrac{1}{{n.\\left( {n + 1} \\right)}} - \\dfrac{1}{{\\left( {n + 1} \\right)\\left( {n + 2} \\right)}} = \\dfrac{2}{{n\\left( {n + 1} \\right)\\left( {n + 2} \\right)}}\\)</p>\r\n<p>\\( \\Rightarrow F = \\dfrac{1}{{1.2.3}} + \\dfrac{1}{{2.3.4}} + \\dfrac{1}{{3.4.5}} +  \\ldots  + \\dfrac{1}{{18.19.20}}\\)</p>\r\n<p>\\( \\Rightarrow 2F = \\dfrac{2}{{1.2.3}} + \\dfrac{2}{{2.3.4}} + \\dfrac{2}{{3.4.5}} +  \\ldots  + \\dfrac{2}{{18.19.20}}\\)</p>\r\n<p>     \\(\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{1}{{1.2}} - \\dfrac{1}{{2.3}} + \\dfrac{1}{{2.3}} - \\dfrac{1}{{3.4}} + \\dfrac{1}{{3.4}} - \\dfrac{1}{{4.5}} +  \\ldots  + \\dfrac{1}{{18.19}} - \\dfrac{1}{{19.20}}\\)</p>\r\n<p>     \\(\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{1}{{1.2}} - \\dfrac{1}{{19.20}} = \\dfrac{{19.10 - 1}}{{19.20}} = \\dfrac{{190 - 1}}{{380}} = \\dfrac{{189}}{{380}}\\)</p>\r\n<p>\\( \\Rightarrow F = \\dfrac{{189}}{{380}}:2 = \\dfrac{{189}}{{380}} \\cdot \\dfrac{1}{2} = \\dfrac{{189}}{{760}}\\)</p>\r\n<p>Vậy \\(F = \\dfrac{{189}}{{760}} \\cdot \\)</p>\r\n<p><strong>Bài 2: </strong></p>\r\n<p>Tính các tổng sau:</p>\r\n<p>a) \\(A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)<strong>                                                                                       </strong></p>\r\n<p>b) \\(B = 1 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{1}{{16}} + \\dfrac{1}{{32}} +  \\ldots  + \\dfrac{1}{{2048}}\\)<strong></strong></p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>Xét các phân số có tử bằng nhau và có mẫu là lũy thừa tăng dần của cùng 1 cơ số thì ta nhân cả 2 vế với đúng cơ số đó. Trường hợp tổng quát:</p>\r\n<p>\\(A = \\dfrac{1}{{{a^1}}} + \\dfrac{1}{{{a^2}}} + \\dfrac{1}{{{a^3}}} +  \\ldots  + \\dfrac{1}{{{a^n}}}\\)\\( \\Rightarrow A.a = a\\left( {\\dfrac{1}{{{a^1}}} + \\dfrac{1}{{{a^2}}} + \\dfrac{1}{{{a^3}}} +  \\ldots  + \\dfrac{1}{{{a^n}}}} \\right)\\)\\( = 1 + \\dfrac{1}{a} +  \\ldots  + \\dfrac{1}{{{a^{n - 1}}}}\\)</p>\r\n<p> </p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>a) \\(A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2A = 2 \\cdot \\left( {\\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}} \\right)\\)</p>\r\n<p>\\( \\Rightarrow 2A = 2 \\cdot \\dfrac{1}{2} + 2 \\cdot \\dfrac{1}{{{2^2}}} + 2 \\cdot \\dfrac{1}{{{2^3}}} + 2 \\cdot \\dfrac{1}{{{2^4}}} +  \\ldots  + 2 \\cdot \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2A = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{2019}}}}\\)</p>\r\n<p>\\(\\,\\,\\,\\,\\,\\,\\,\\,\\,A = \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2A - A = \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{2019}}}}} \\right) - \\left( {\\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{2020}}}}} \\right)\\)</p>\r\n<p>\\( \\Rightarrow A = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{2019}}}} - \\dfrac{1}{2} - \\dfrac{1}{{{2^2}}} - \\dfrac{1}{{{2^3}}} - \\dfrac{1}{{{2^4}}} -  \\ldots  - \\dfrac{1}{{{2^{2020}}}}\\)</p>\r\n<p>\\( \\Rightarrow A = 1 - \\dfrac{1}{{{2^{2020}}}} = \\dfrac{{{2^{2020}} - 1}}{{{2^{2020}}}}\\)</p>\r\n<p>Vậy \\(A = \\dfrac{{{2^{2020}} - 1}}{{{2^{2020}}}}\\).</p>\r\n<p>b) \\(B = 1 + \\dfrac{1}{2} + \\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{1}{{16}} +  \\ldots  + \\dfrac{1}{{2048}}\\) \\( = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{11}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B = 2 \\cdot \\left( {1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{11}}}}} \\right)\\)</p>\r\n<p>\\( \\Rightarrow 2B = 2.1 + 2 \\cdot \\dfrac{1}{2} + 2 \\cdot \\dfrac{1}{{{2^2}}} + 2 \\cdot \\dfrac{1}{{{2^3}}} + 2 \\cdot \\dfrac{1}{{{2^4}}} +  \\ldots  + 2 \\cdot \\dfrac{1}{{{2^{11}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B = 2 + 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{10}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B = 3 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} +  \\ldots  + \\dfrac{1}{{{2^{10}}}}\\)</p>\r\n<p>\\(\\,\\,\\,\\,\\,\\,\\,\\,\\,B = 1 + \\dfrac{1}{2} + \\dfrac{1}{{{2^2}}} + \\dfrac{1}{{{2^3}}} + \\dfrac{1}{{{2^4}}} +  \\ldots  + \\dfrac{1}{{{2^{11}}}}\\)</p>\r\n<p>\\( \\Rightarrow 2B - B = 2 - \\dfrac{1}{{{2^{11}}}}\\)\\( \\Rightarrow B = \\dfrac{{{2^{12}} - 1}}{{{2^{11}}}}\\);</p>\r\n<p>Vậy \\(B = \\dfrac{{{2^{12}} - 1}}{{{2^{11}}}} \\cdot \\)</p>\r\n<p><strong>Bài 3:</strong></p>\r\n<p>a) Tính tổng sau: \\(A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\)</p>\r\n<p>b) Chứng minh rằng biểu thức \\(B\\) có giá trị bằng \\(\\dfrac{1}{2}\\) với \\(B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}.\\)</p>\r\n<p><strong>Phương pháp</strong></p>\r\n<p>+) Áp dụng quy tắc dấu ngoặc, nhóm các hạng tử.</p>\r\n<p>+) Áp dụng công thức tính tổng của 1 dãy các số tự nhiên liên tiếp: \\(1 + 2 +  \\ldots  + n = \\dfrac{{n + 1}}{2} \\cdot n = \\dfrac{{n.\\left( {n + 1} \\right)}}{2}\\)</p>\r\n<p><strong>Lời giải</strong></p>\r\n<p>a) \\(A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\)</p>\r\n<p>Ta có:</p>\r\n<p>\\(\\begin{array}{l}A = \\dfrac{{1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\\\\\,\\,\\,\\,\\, = \\dfrac{{1 + 1 + 2 + 1 + 2 + 3 +  \\ldots  + 1 + 2 + 3 +  \\ldots 2020}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\\\\\,\\,\\,\\,\\, = \\dfrac{{\\left( {1 + 1 + 1 +  \\ldots  + 1} \\right) + \\left( {2 + 2 +  \\ldots 2} \\right) + \\left( {3 + 3 + 3 + 3 \\ldots } \\right) +  \\ldots  + 2020}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\end{array}\\)<strong></strong></p>\r\n<p>\\(\\begin{array}{l}\\,\\,\\,\\,\\, = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}\\\\\\,\\,\\,\\,\\, = 1\\end{array}\\)<strong></strong></p>\r\n<p>b) Chứng minh rằng biểu thức \\(B\\) có giá trị bằng \\(\\dfrac{1}{2}\\) với \\(B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}.\\)</p>\r\n<p>Với biểu thức \\(B\\), xét tử số ta có:</p>\r\n<p>\\(\\,\\,\\,\\,1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1\\)</p>\r\n<p>\\( = 1 + \\left( {1 + 2} \\right) + \\left( {1 + 2 + 3} \\right) +  \\ldots  + \\left( {1 + 2 + 3 +  \\ldots  + 2020} \\right)\\)</p>\r\n<p>\\( = \\dfrac{{0 + 1}}{2} \\cdot 2 + \\dfrac{{1 + 2}}{2} \\cdot 2 + \\dfrac{{1 + 3}}{2} \\cdot 3 +  \\ldots  + \\dfrac{{1 + 2020}}{2} \\cdot 2020\\)</p>\r\n<p>\\( = \\dfrac{1}{2} \\cdot 2 + \\dfrac{3}{2} \\cdot 2 + \\dfrac{4}{2} \\cdot 3 +  \\ldots  + \\dfrac{{2021}}{2} \\cdot 2020\\)</p>\r\n<p>\\( = \\dfrac{{1.2}}{2} + \\dfrac{{2.3}}{2} + \\dfrac{{3.4}}{2} +  \\ldots  + \\dfrac{{2020.2021}}{2}\\)</p>\r\n<p>\\( = \\dfrac{1}{2} \\cdot \\left( {1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021} \\right)\\)</p>\r\n<p>\\(\\begin{array}{l} \\Rightarrow B = \\dfrac{{1.2020 + 2.2019 + 3.2018 +  \\ldots  + 2020.1}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\dfrac{{\\dfrac{1}{2} \\cdot \\left( {1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021} \\right)}}{{1.2 + 2.3 + 3.4 +  \\ldots  + 2020.2021}} = \\dfrac{1}{2}.\\end{array}\\)</p>\r\n<p>Vậy \\(B = \\dfrac{1}{2} \\cdot \\)</p>","youtube_id":"","video_start_time":"0"}],"comment_url":"https://loigiaihay.com/comment/dang-6-day-phan-so-viet-theo-quy-luat-chu-de-6-on-he-toan-6-c941a110990.html","origin_url":"https://loigiaihay.com/dang-6-day-phan-so-viet-theo-quy-luat-chu-de-6-on-he-toan-6-c941a110990.html","relate_link":[]},"callouts":[],"relate_articles":[],"others_in_cat":[{"article_id":111045,"cat_id":941,"publish_time":1655967180,"is_link":0,"thumbnail":"","title":"Dạng 7. Hai bài toán về phân số Chủ đề 6 Ôn hè Toán 6","alias":"dang-7-hai-bai-toan-ve-phan-so-chu-de-6-on-he-toan-6","introtext":"a) Tìm a/b của m là ","display_type":2,"article_type":1},{"article_id":110987,"cat_id":941,"publish_time":1655887800,"is_link":0,"thumbnail":"","title":"Dạng 5. Tìm x Chủ đề 6 Ôn hè Toán 6","alias":"dang-5-tim-x-chu-de-6-on-he-toan-6","introtext":"Dùng quy tắc thực hiện phép tính, quy tắc chuyển vế, quy tắc dấu ngoặc để đưa về các dạng quen thuộc để tìm x:","display_type":2,"article_type":1},{"article_id":110939,"cat_id":941,"publish_time":1655804340,"is_link":0,"thumbnail":"","title":"Dạng 4. Tính bằng cách hợp lí Chủ đề 6 Ôn hè Toán 6","alias":"dang-4-tinh-bang-cach-hop-li-chu-de-6-on-he-toan-6","introtext":"Sử dụng các tính chất của phép cộng, phép nhân phân số:\r\n+) Phép cộng:","display_type":2,"article_type":1},{"article_id":110659,"cat_id":941,"publish_time":1655347080,"is_link":0,"thumbnail":"","title":"Dạng 3. Thực hiện phép tính Chủ đề 6 Ôn hè Toán 6","alias":"dang-3-thuc-hien-phep-tinh-chu-de-6-on-he-toan-6","introtext":"1. Phép cộng\r\n\r\nMuốn cộng hai phân số cùng mẫu, ta cộng các tử và giữ nguyên mẫu.","display_type":2,"article_type":1},{"article_id":110658,"cat_id":941,"publish_time":1655346720,"is_link":0,"thumbnail":"","title":"Dạng 2. So sánh phân số Chủ đề 6 Ôn hè Toán 6","alias":"dang-2-so-sanh-phan-so-chu-de-6-on-he-toan-6","introtext":"Tính chất: Nếu a < b thì -a > -b\r\n\r\nCách 1: Đưa về 2 phân số có cùng mẫu số","display_type":2,"article_type":1},{"article_id":110039,"cat_id":941,"publish_time":1654766580,"is_link":0,"thumbnail":"","title":"Dạng 1. Quy đồng mẫu số các phân số Chủ đề 6 Ôn hè Toán 6","alias":"dang-1-quy-dong-mau-so-cac-phan-so-chu-de-6-on-he-toan-6","introtext":"Để quy đồng nhiều phân số, ta thường làm như sau:","display_type":2,"article_type":1}],"event_info":{"title":"Ôn tập hè Chủ đề 6. Phân số. Các bài toán về phân số","cat_id":941,"alias":"on-tap-he-chu-de-6-phan-so-cac-bai-toan-ve-phan-so","introtext":"","type":"event","item_id":26483,"item_type":2},"banner":[],"top_banner":"","content_expanded":[],"child_events":[{"id":"26483","title":"Ôn tập hè Chủ đề 6. Phân số. Các bài toán về phân số","short_title":"Chủ đề 6. Phân số. Các bài toán về phân số","alias":"on-tap-he-chu-de-6-phan-so-cac-bai-toan-ve-phan-so","picture":"","is_link":"0","links":"","is_hot":"0","is_all":"0","category_id":"941","ordering":"26483","parent_id":"0"}],"parent_events":[],"cfg_info":{"last_cfg_time":"1521866679","ad.popup_frequency":"0","ad.adsense.detail_bottom.unit_id":"ca-app-pub-8529835372050931/9927553003","ad.adsense.popup.unit_id":"ca-app-pub-8529835372050931/2161779402","current.force.update.version":"181","support.email":"loigiaihay.com@gmail.com","support.fanpage":"https://www.facebook.com/Loigiaihay/","popup_rating_after":"86400","admob.app_id":"ca-app-pub-8529835372050931~4358271409","admob.unit.article_detail_bottom":"ca-app-pub-8529835372050931/1883637630","admob.unit.article_detail_top":"ca-app-pub-8529835372050931/8604898230","admob.unit.event_article_top":"ca-app-pub-8529835372050931/7406887709","admob.unit.interestial":"ca-app-pub-8529835372050931/2741937409","ios_appstore_link":"https://itunes.apple.com/us/app/loigiaihay/id1209891610?mt=8","android_appstore_link":"https://play.google.com/store/apps/details?id=com.tp.loigiaihay","min_version":"2.1.3","latest_version":"1.3.9","force_update":false},"popup_ads":[]}