Practice section 14 of 17

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]

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]

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]

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], and (b) $\sum_{n=50}^{100}1/(\sqrt{n+1}+\sqrt{n-1})$. [1]

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

Given $\sum_{r=2}^{n-1}[2r+4(1/(r+1)-1/r)]=n^2-n+4/n-4$, show

[3]

Given $A_n=\sum_{r=1}^n1/[r(r+2)]=3/4-1/[2(n+1)]-1/[2(n+2)]$:

(i) Deduce $A_n<3/4$ for all positive integers $n$. [1]

(ii) Given also $\sum_{r=1}^n1/[r(r+1)]=1-1/(n+1)$, find $\sum_{r=1}^n1/[2r(2r+2)]$. [2]

(iii) Use these results to find $\sum_{r=1}^n1/(4r^2-1)$ in terms of $n$. [4]