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

Solution:

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

$a_{n}=n(n+1)(n+4)$

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

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

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+4))$

$=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}+5 \mathrm{n}^{2}+4 \mathrm{n}\right)$

$=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{3}\right)+5 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)+4 \sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})$

Note:

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

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

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

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

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,

$S_{n}=\sum_{n=1}^{n}\left(n^{3}\right)+5 \sum_{n=1}^{n}\left(n^{2}\right)+4 \sum_{n=1}^{n}(n)$

From the above identities,

$S_{n}=\sum_{n=1}^{n}\left(n^{3}\right)+5 \sum_{n=1}^{n}\left(n^{2}\right)+4 \sum_{n=1}^{n}(n)$

$S_{n}=\left(\frac{n(n+1)}{2}\right)^{2}+5\left(\frac{n(n+1)(2 n+1)}{6}\right)+4\left(\frac{n(n+1)}{2}\right)$

$S_{n}=\frac{(n(n+1))^{2}}{4}+5\left(\frac{n(n+1)(2 n+1)}{6}\right)+2(n(n+1))$

$=n(n+1)\left(\frac{n(n+1)}{4}+5\left(\frac{2 n+1}{6}\right)+2\right)$

$=\frac{n(n+1)}{24}(6 n(n+1)+20(2 n+1)+48)$

$=\frac{n(n+1)}{24}\left(6 n^{2}+46 n+68\right)$

$=\frac{n(n+1)}{12}(63+23 n+34)$

Hence, the Sum of the series, $S_{n}=\frac{n(n+1)}{12}(63+23 n+34)$