Find the sum of the series whose nth term is given by

Solution:

It is given in the question that the $\mathrm{n}^{\text {th }}$ term of the series,

$a_{n}=4 n^{3}+6 n^{2}+2 n$

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

$S_{n}=\sum_{n=1}^{n} a_{n}$

$S_{n}=\sum_{n=1}^{n}\left(4 n^{3}+6 n^{2}+2 n\right)$

$=\sum_{n=1}^{n}\left(4 n^{3}\right)+\sum_{n=1}^{n}\left(6 n^{2}\right)+\sum_{n=1}^{n}(2 n)$

$=4 \sum_{n=1}^{n}\left(n^{3}\right)+6 \sum_{n=1}^{n}\left(n^{2}\right)+2 \sum_{n=1}^{n}(n)$

Note:

I. Sum of first $n$ natural numbers, $1+2+3+\ldots n$,

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

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, for the given series, we need to find,

$\mathrm{S}_{\mathrm{n}}=4 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}\right)+6 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+2 \sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$

From the above identities,

$\mathrm{S}_{\mathrm{n}}=4 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}\right)+6 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+2 \sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$

$\mathrm{S}_{\mathrm{n}}=4\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}+6\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right)+2\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)$

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

$=n(n+1)[n(n+1)(2 n+1)+1]$

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

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

Hence, the Sum of the series, $S_{n}=n(n+1)^{2}(n+2)$