Show that the sequence, for satisfies the recurrence relation , for all integers. Question 11: The sequence is given by the formula dn = 2 n 1 for n 0 – Substituting k 1 for n we get dk-1 = 2k-1 1 – – Substituting k 2 for n we get dk-2 = 2k-2 1 – – We want to prove that dk = 3dk-1 2dk-2 – R.H.S. = 3(2k 1 1) 2(2k 2 1) – – – – – = 3 · 2k 1 3 2 · 2k 2 + 2 – – – – = 3 · 2k 1 2k 1 1 – – – – = (3 1) · 2k 1 1 – – – = 2 · 2k 1 1 = 2k 1 = dk = L.H.S. – – – Question 12: Define a sequence by the formula , for all integers. Show that this sequence satisfies the recurrence relation , for all integers. Similarly solve the questions as 11.
Question 11: Show that the sequence, for satisfies the recurrence relation , for all integers. The sequence is given by the formula dn = 2 n – 1 for n 0 Substituting k – 1 for n we get dk-1 = 2k-1 – 1 Substituting k – 2 for n we get dk-2 = 2k-2 – 1 We want to prove that dk = 3dk-1 – 2dk-2 R.H.S. = 3(2k – 1 – 1) – 2(2k – 2 – 1) = 3 · 2k – 1 – 3 – 2 · 2k – 2 + 2 = 3 · 2k – 1 – 2k – 1 – 1 = (3 – 1) · 2k – 1 – 1 = 2 · 2k – 1 – 1 = 2k – 1 = dk = L.H.S. Question 12: Define a sequence by the formula , for all integers. Show that this sequence satisfies the recurrence relation , for all integers. Similarly solve the questions as 11.
Solution%20of%20Practice%20questions%20Sequence%20and%20Series%20%20(19%20-22)%20Updated.docx