Review practice · Section 1 of 4

Evaluate $\displaystyle\sum_{r=2}^{n}(2^{-r}+2nr+n^2)$, giving your answer in terms of $n$.4 marks

Given $\sum_{r=1}^n(3r^2+3r+1)=(n+1)^3-1$, show $\sum_{r=1}^nr^2=\tfrac16n(n+1)(2n+1)$.4 marks

Given $\sum_{r=1}^n(2r+3)^2=\tfrac n3(4n^2+24n+47)$, find $5^2+7^2+9^2+\cdots+2011^2$.2 marks

Given $\sum_{r=0}^n1/[(r+1)(r+2)(r+3)]=\tfrac14-1/[2(n+2)(n+3)]$, deduce $\sum_{r=5}^{20}1/[(r+1)(r+2)(r+3)]$.3 marks

Given $\sum_{r=1}^nr^2=n(n+1)(2n+1)/6$, find the sum of the first $n$ terms of $(1)(2)+(3)(4)+(5)(6)+\cdots$.3 marks

Given $\sum_{n=1}^N(\sqrt{n+1}-\sqrt{n-1})=\sqrt N+\sqrt{N+1}-1$:

(i) Find exactly (a) $\sum_{n=50}^{100}(\sqrt{n+1}-\sqrt{n-1})$ 2 marks, and (b) $\sum_{n=50}^{100}1/(\sqrt{n+1}+\sqrt{n-1})$.1 mark

(ii) State with a reason whether $\sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n-1})$ converges.1 mark