It is given that $\sum\limits_{r=1}^{n}\dfrac{2r+1}{r^2(r+1)^2}=1-\dfrac{1}{(n+1)^2}$ for all

Question

It is given that $\sum\limits_{r=1}^{n}\dfrac{2r+1}{r^2(r+1)^2}=1-\dfrac{1}{(n+1)^2}$ for all $n\in\mathbb Z^+$. Find $\sum\limits_{r=5}^{2n}\dfrac{2r+1}{r^2(r+1)^2}$ in terms of $n$.[3] Find $\sum\limits_{r=2}^{n}\dfrac{2r-1}{(r-1)^2r^2}$, leaving your answer in the form of $a-\dfrac{b}{n^2}$, where $a$ and $b$ are constants to be determined. Hence find $\sum\limits_{r=2}^{\infty}\dfrac{2r-1}{(r-1)^2r^2}$.[3]

Accepted Answer

(a) $\dfrac1{25}-\dfrac1{(2n+1)^2}$. (b) $1-\dfrac1{n^2}$, $a=b=1$; infinite sum $1$.

2026 CJC Promo Q5

2026 CJC Promo Q5

Solution:

Solution locked