<s>[INST] Find $S_1^2+S_2^2+S_3^2+...+S_{2n-1}^2$ Question: If $S_1,S_2,S_3,...,S_n$ be the sums of $n$ infinite G.P. series respectively whose first terms are respectively $1,2,3,...,n$ and common ratios $\frac{1}{2},\frac{1}{3},\frac{1}{4},...,\frac{1}{n+1}$ respectively, then find the value of $S_1^2+S_2^2+S_3^2+...+S_{2n-1}^2$.
My attempt:
$S_k=\frac{k}{1-\frac{1}{k+1}}$
$\implies S_k=k+1$
$\implies S_k^2=(k+1)^2$
$\implies S=\sum\limits_{k=1}^{2n-1}S_k^2$
$\implies S=\sum\limits_{k=1}^{2n-1}(k+1)^2$
$\implies S=\sum\limits_{k=1}^{2n-1}(k^2+1+2k)$
$\implies S=\sum\limits_{k=1}^{2n-1}(k^2)+\sum\limits_{k=1}^{2n-1}(1)+\sum\limits_{k=1}^{2n-1}(2k)$
$\implies S=\frac{(2n-1)(2n)[2(2n-1)+1]}{6}\ \ \ \ \ \ \ +(2n-1)+\frac{2(2n-1)(2n)}{2}$
$\implies S=\frac{(4n^2-2n)(4n-1)}{6}\;\;\;+2n-1+4n^2-2n$
$\implies S=\frac{16n^3-4n^2-8n^2+2n}{6}\;\;+2n-1+4n^2-2n$
$\implies S=\frac{16n^3-4n^2-8n^2+2n-6+24n^2}{6}$
$\implies S=\frac{16n^3+12n^2+2n-6}{6}$
$\implies S=\frac{8n^3+6n^2+n-3}{3}$
My problem: The correct answer is $\frac{n(2n+1)(4n+1)}{3}$ which on simplification gives $\frac{8n^3+6n^2+n}{3}$. Where have i gone wrong? Please help.
 [/INST] Since
$$
\begin{align}
(k+1)^2
&=(k+1)k+(k+1)\\
&=2\binom{k+1}{2}+\binom{k+1}{1}
\end{align}
$$
we have
$$
\begin{align}
\sum_k\binom{2n-1-k}{2n-1-k}\left[2\binom{k+1}{k-1}+\binom{k+1}{k}\right]
&=2\binom{2n+1}{2n-2}+\binom{2n+1}{2n-1}\\
&=\frac{8n^3+6n^2+n}3
\end{align}
$$
Therefore, leaving out the $k=0$ term, we get
$$
\sum_{k=1}^{2n-1}(k+1)^2=\bbox[5px,border:2px solid #C0A000]{\frac{8n^3+6n^2+n-3}3}
$$
</s>