Prove that the sum to n terms of the Sequence:
$1^2/(1×3),2^2/(3×5),3^2/(5×7),...$ is $ n(n+1)/2(2n+1).$
Im having trouble with this question, firstly ive begun by stating that p(n) denotes the statement.
Then to prove that p(1) holds where $n = 1$ $n(n+1)2(2n+1).$ = 1/3 and $1^2/1*3$ = $1/3$ therefore true
Now i know I have to show $n=k$ and therefore $n=k+1$ but how can i do this? I'm very tired so clear instruction would be great.
We need to show that for every $n$, $$\sum_{k=1}^n\frac{k^2}{(2k-1)(2k+1)}=\frac{n(n+1)}{2(2n+1)}$$
For $n=1$, you already checked (both sides give $1/3$). Now, suppose that the statement holds for some $n$ (that is, $p(n)$ is valid). Then
$$\sum_{k=1}^{n+1}\frac{k^1}{(2k-1)(2k+1)}=\left(\sum_{k=1}^{n+1}\frac{k^1}{(2k-1)(2k+1)}\right)+\frac{(n+1)^2}{(2(n+1)-1)(2(n+1)+1)}=\frac{n(n+1)}{2(2n+1)}+\frac{(n+1)^2}{(2n+1)(2n+3)}$$
You can work a little on this and obtain $\frac{(n+1)(n+2)}{2(2n+3)}$ (hint:$2n^2+5n+2=(2n+1)(n+2)$), which is equal to $\frac{(n+1)((n+1)+1)}{2(2(n+1)+3)}$ (knowing where you should get to is a big help). Then $$\sum_{k=1}^{n+1}\frac{k^2}{(2k-1)(2k+1)}=\frac{(n+1)((n+1)+1)}{2(2(n+1)+1)},$$ which says that the statement is valid for $n+1$ as well. By induction, this is valid for all $n$.