# Find the sum of the series:

Question:

Find the sum of the series:

$\left(3 \times 1^{2}\right)+\left(5 \times 2^{2}\right)+\left(7 \times 3^{2}\right)+\ldots$ to $n$ terms

Solution:

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

For that, first, we need to find the $\mathrm{n}^{\text {th }}$ 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 $\left(3 \times 1^{2}\right)+\left(5 \times 2^{2}\right)+\left(7 \times 3^{2}\right)+\ldots$ to $n$ terms.

The series can be written as, $\left[\left(3 \times 1^{2}\right),\left(5 \times 2^{2} \ldots\left((2 n+1) \times n^{2}\right]\right.\right.$.

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

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

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

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

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

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

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

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

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

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

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,

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

From, the above identities,

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

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

$\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)\left[\frac{3 \mathrm{n}^{2}+5 \mathrm{n}+1}{3}\right]$

So, Sum of the series, $S_{n}=\left(\frac{n(n+1)}{6}\right)\left(3 n^{2}+5 n+1\right)$