nth Finally the terms can be completed as shown. Extending the table backwards will a, b c allow us to find the values of and nth an2 bn c. in the term + + a 2a = 2 = 1 → a b b + = 4 = 3 → nth C1 for correctly finding the term c = –2 10 nth n2 Hence term is + 3n 2. – 52 n2)2 M1 2 M1 choosing two smallest terms, squaring and adding M (m2 + (2mn)2 = – 3 2m2n2+ + 4m2n2 A1 A1 for cao m4 – n4 m4 n4 A1 A1 for cao = + 2m2n2 + C1 C1 for showing the factorisation leads to the given n2)2 = (m2 + result 4 53 Because the differences between C3 2 C1 for explaining the about the differences of the terms M consecutive terms are 2, 3, 4, 5, 6, etc. being the set of integers (even and odd alternately); when generating the triangular number C1 for explaining the pattern is odd, even, odd, even sequence starting with the odd 1, add and so on even to odd to generate odd add …3; odd to odd to generate even add …6; even to even to generate even …10. Add odd to even to generate odd …15. We are now back again where we add even to odd to generate odd, and the whole sequence continues in the same C1 for explaining the complete sequence of combining way, continually giving two odd, two odd and even to generate the final sequence even, etc. 3
Finally the nth terms can be completed as shown. Extending the table backwards will allow us to find the values of a, b and c in the nth term an2 + bn + c. 2a = 2 → a = 1 a + b = 4 → b = 3 c = –2 Hence nth term is n2 + 3n – 2. C1 for correctly finding the nth term 10 52 (m2 – n2)2 + (2mn)2 = m4 – 2m2n2+ n4 + 4m2n2 = m4 + 2m2n2 + n4 = (m2 + n2)2 M1 A1 A1 C1 2 3 M1 choosing two smallest terms, squaring and adding A1 for cao A1 for cao C1 for showing the factorisation leads to the given result M 4 53 Because the differences between consecutive terms are 2, 3, 4, 5, 6, etc. (even and odd alternately); when generating the triangular number sequence starting with the odd 1, add even to odd to generate odd …3; add odd to odd to generate even …6; add even to even to generate even …10. Add odd to even to generate odd …15. We are now back again where we add even to odd to generate odd, and the whole sequence continues in the same way, continually giving two odd, two even, etc. C3 2 C1 for explaining the about the differences of the terms being the set of integers C1 for explaining the pattern is odd, even, odd, even and so on C1 for explaining the complete sequence of combining odd and even to generate the final sequence M 3
Higher_Skills_Algebra_MarkScheme_Edexcel.docx