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 $n^{\text {th }}$ 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 $3,15,35,63 \ldots$ to $n$ terms.

The series can be written as, $\left[2^{2}-1,4^{2}-1,6^{2}-1 \ldots(2 n)^{2}-1\right]$.

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

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

$a_{n}=4 n^{2}-1$

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

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

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

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

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

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

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

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,

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

From, the above identities,

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

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

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

$S_{n}=\left(\frac{n}{3}\right)\left[4 n^{2}+6 n-1\right]$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{3}\left[4 \mathrm{n}^{2}+6 \mathrm{n}-1\right]$