Question 0507

Sigma Notation
2007 Paper 2 Question 2 Variant

Question

A sequence u1,u2,u3,…u_1, \allowbreak u_2, \allowbreak u_3, \ldots is such that
un+1=un−3n2−3n+1(n−1)3n3,for all n≥1,u_{n+1} = u_n - \frac{3 n^2 - 3 n + 1}{(n - 1)^3n^3}, \quad \textrm{for all } n \geq 1,
and un=1(n−1)3.{\displaystyle u_n = \frac{1}{(n - 1)^3}.}
(i)
Find ∑n=4N3n2−3n+1(n−1)3n3.{\displaystyle \sum_{n=4}^N \frac{3 n^2 - 3 n + 1}{(n - 1)^3n^3}.}
[2]
(ii)
Give a reason why the series in part (i) is convergent and state the sum to infinity.
[2]
(iii)
Use your answer to part (i) to find ∑n=5N3n2−9n+7(n−2)3(n−1)3.{\displaystyle \sum_{n=5}^N \frac{3 n^2 - 9 n + 7}{(n - 2)^3(n - 1)^3}.}
[2]

Answer

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