Question 0509

Sigma Notation
2009 Paper 1 Question 1 Variant

Question

(a)
Show that
2n1+1n3n+1=An+Bn3n,\frac{ 2 }{ n - 1 } + \frac{ 1 }{ 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=5n5n1r3r.{\displaystyle \sum_{r=5}^n \frac{5 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=55n1r3r{\displaystyle \sum_{r=5}^\infty \frac{5 n - 1}{r^3-r}} converges, and write down its value.
[2]

Answer