Question

If $S_k=\frac{1+2+3+4+...+k}{k}$, find the value of $S_1^2+S_2^2+S_3^2+...+S_n^2.$

Also, determine $\sum \left(\frac{S_n}{n}\right).$

Answer 1

Given, $S_k=\frac{1+2+3+...+k}{k}=\frac{k(k+1)}{2k}=\frac{k+1}{2}[\because \sum n=\frac{n(n+1)}{2}]$

$\therefore S_1^2+S_2^2+S_3^2+...+S_n^2={\sum }_{k=1}^n{S}_k^2={\sum }_{k=1}^n{\left(\frac{k+1}{2}\right)}^2=\frac{1}{4}{\sum }_{k=1}^n(K+1{)}^2=\frac{1}{4}{\sum }_{k=1}^n(k^2+2k+1)$

= $\frac{1}{4}[{\sum }_{k=1}^n{k}^2+2{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}1]=\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}+n]$

= $\frac{n}{4}[\frac{(n+1)(2n+1)}{6}+n+1+1]=\frac{n}{4}\left(\frac{2n^2+3n+1+6n+12}{6}\right)$

= $\frac{n}{24}(2n^2+9n+13)$

Now, $\sum \frac{S_n}{n}=\sum \frac{1}{24}(2n^2+9n+13)=\frac{1}{24}(2\sum n^2+9\sum n+13\sum 1)$

= $\frac{1}{24}[2\times \frac{n(n+1)(2n+1)}{6}+9\times \frac{n(n+1)}{2}+13n]$

= $\frac{n}{24}[2\times \frac{(2n^2+3n+1)}{6}+\frac{9n+9}{2}+13]$

= $\frac{n}{24}\left(\frac{4n^2+6n+2+27n+27+78}{6}\right)$

= $\frac{n}{144}(4n^2+33n+107)$