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 $\mathrm{n}^{\text {th }}$ 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 $\left(1 \times 2^{2}\right)+\left(3 \times 3^{2}\right)+\left(5 \times 4^{2}\right)+\ldots$ to $n$ terms.

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

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

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

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

$=2 n^{3}+3 n^{2}-1$

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

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

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

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,

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

From, the above identities,

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

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

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

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

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

$\mathrm{S}_{\mathrm{n}}=\left(\frac{\mathrm{n}}{2}\right)\left[\mathrm{n}^{3}+4 \mathrm{n}^{2}+4 \mathrm{n}-1\right]$

So, Sum of the series, $S_{n}=\left(\frac{n}{2}\right)\left[n^{3}+4 n^{2}+4 n-1\right]$