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,

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

Now, we need to find the sum of this series, $S_{n}$.

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

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

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

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

Note:

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

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

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,

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

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

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

From the above identities

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

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

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

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

$=\frac{\mathrm{n}}{2}(\mathrm{n}+1)(2 \mathrm{n}+3)$

Hence, Sum of the series, $S_{n}=\frac{n}{2}(n+1)(2 n+3)$