Find $S_1^2+S_2^2+S_3^2+...+S_{2n-1}^2$

Question

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.

Answer 1

$$\begin{align} \sum_{r=1}^{2n-1}S_r^2&=\sum_{r=1}^{2n-1}\left(\frac r{1-\frac 1{r+1}}\right)^2\\ &=\sum_{r=1}^{2n-1}(r+1)^2\\ &=\sum_{r=2}^{2n}r^2\\ &=\frac 16 (2n)(2n+1)(4n+1)-1\\ &=\boxed{\frac {n(2n+1)(4n+1)}3}-1\\ &=\frac {8n^3+6n^2+n-3}3\\ &=\frac {(2n-1)(4n^2+5n+3)}3\end{align}$$

Answer 2

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 $$ \sum_{k=1}^{2n-1}(k+1)^2=\bbox[5px,border:2px solid #C0A000]{\frac{8n^3+6n^2+n-3}3} $$