Question 0509

Sigma Notation
2009 Paper 1 Question 1 Variant

Question

(a)
Show that
1n−1+2n−3n+1=An+Bn3−n,\frac{ 1 }{ n - 1 } + \frac{ 2 }{ n } - \frac{ 3 }{ n + 1 } = \frac{An+B}{n^3-n},
where A{A} and B{B} are constants to be found.
[2]
(b)
Hence find ∑r=2n2n−1r3−r.{\displaystyle \sum_{r=2}^n \frac{2 n - 1}{r^3-r}.}
(There is no need to express your answer as a single algebraic fraction.)
[3]
(c)
Give a reason why the series ∑r=2∞2n−1r3−r{\displaystyle \sum_{r=2}^\infty \frac{2 n - 1}{r^3-r}} converges, and write down its value.
[2]

Answer

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