# Solve this

Question:

If $\mathrm{S}_{\mathrm{k}}=\frac{(1+2+3+\ldots \mathrm{k})}{\mathrm{k}}$ prove that ,

$\left(\mathrm{S}_{1}^{2}+\mathrm{S}_{2}^{2}+\cdots \mathrm{S}_{\mathrm{n}}^{2}\right)=\frac{\mathrm{n}}{24}\left(2 \mathrm{n}^{2}+9 \mathrm{n}+13\right)$

Solution:

Given, $\mathrm{S}_{\mathrm{k}}=\frac{(1+2+3+\cdots \mathrm{k})}{\mathrm{k}}$

To prove: $\left(\mathrm{S}_{1}{ }^{2}+\mathrm{S}_{2}{ }^{2}+\cdots \mathrm{S}_{\mathrm{n}}{ }^{2}\right)=\frac{\mathrm{n}}{24}\left(2 \mathrm{n}^{2}+9 \mathrm{n}+13\right)$

$\left(\mathrm{S}_{\mathrm{k}}\right)=\frac{1+2+3+\cdots \mathrm{k}}{\mathrm{k}}$

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$

II. Sum of squares of first $n$ natural numbers, $1^{2}+2^{2}+3^{2}+\ldots n^{2}$,

$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}$

III. Sum of cubes of first $n$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots . n^{3}$,

$\sum_{k=1}^{n} k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$

IV. Sum of a constant k, N times,

$\sum_{\mathrm{k}=1}^{\mathrm{N}} \mathrm{k}=\mathrm{Nk}$

So

$\left(\mathrm{S}_{\mathrm{k}}\right)=\frac{\mathrm{k}(\mathrm{k}+1)}{2 \mathrm{k}}=\frac{\mathrm{k}+1}{2}$

Now, the Left hand side of the condition given in the question can be written as,

$\left(\mathrm{S}_{1}{ }^{2}+\mathrm{S}_{2}{ }^{2}+\cdots \mathrm{S}_{\mathrm{n}}{ }^{2}\right)=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}{ }^{2}$

The required LHS, $S_{n}=\sum_{k=1}^{n} S_{k}{ }^{2}$

$\sum_{k=1}^{n} S_{k}^{2}=\sum_{k=1}^{n}\left(\frac{k+1}{2}\right)^{2}$

$=\sum_{k=1}^{n} \frac{k^{2}+2 k+1}{4}$

$=\frac{1}{4}\left(\sum_{k=1}^{n} k^{2}+2 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1\right)$

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

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

$=\frac{n}{4}\left[\left(\left(\frac{(n+1)(2 n+7)+6}{6}\right)\right)\right]$

$S_{n}=\frac{n}{24}\left[2 n^{2}+9 n+13\right]$

So,

$\left(S_{1}^{2}+S_{2}^{2}+\cdots S_{n}^{2}\right)=\sum_{k=1}^{n} S_{k}^{2}=S_{n}=\frac{n}{24}\left[2 n^{2}+9 n+13\right]$

$\left(S_{1}^{2}+S_{2}^{2}+\cdots S_{n}^{2}\right)=\frac{n}{24}\left[2 n^{2}+9 n+13\right]$

With $S_{k}=\frac{(1+2+3+\cdots k)}{k}$