Find the sum of the series:

Solution:

In the given question we need to find the sum of the series.

For that, first, we need to find the $n^{\text {th }}$ term of the series so that we can use summation of the series with standard identities and get the required sum.

The series given is $\left(1 \times 2^{2}\right)+\left(2 \times 3^{2}\right)+\left(3 \times 4^{2}\right)+\ldots$ to $n$ terms.

The series can be written as, $\left[\left(1 \times(1+1)^{2}\right),\left(2 \times(2+1)^{2} \ldots\left(n \times(n+1)^{2}\right]\right.\right.$.

So, $\mathrm{n}^{\text {th }}$ term of the series,

$a_{n}=n(n+1)^{2}$

$a_{n}=n^{3}+2 n^{2}+n$

Now, we need to find the sum of this series, Sn

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{n}}$

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}+2 \mathrm{n}^{2}+\mathrm{n}\right)$

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 $\mathrm{n}$ natural numbers, $1^{3}+2^{3}+3^{3}+\ldots \ldots \mathrm{n}^{3}$,

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

IV. Sum of a constant k, N times,

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

So, for the given series, we need to find,

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}\right)+2 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$

From, the above identities,

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

$\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}+2\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right)+\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)$

$=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{2(2 \mathrm{n}+1)}{3}+1\right]$

$\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\left[\frac{3 \mathrm{n}^{2}+11 \mathrm{n}+10}{6}\right]$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{12}\right)\left(3 \mathrm{n}^{2}+11 \mathrm{n}+10\right)$