Find the sum of the series:


Find the sum of the series:

$\left(2^{2}+4^{2}+6^{2}+8^{2}+\ldots\right.$ to $n$ terms $)$



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 $\ldots 2^{2}+4^{2}+6^{2}+8^{2}+\ldots$ to $n$ terms.

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

$\left.(2 \times 3)^{2} \ldots(2 \times n)^{2}\right]$

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

$a_{n}=(2 n)^{2}=4 n^{2}$

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

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

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

$=4 \sum_{n=1}^{n}\left(n^{2}\right)$


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

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

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,

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

From, the above identities,

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

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

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

$S_{n}=\frac{2}{3}[n(n+1)(2 n+1)]$

So, Sum of the series, $\mathrm{S}_{\mathrm{n}}=\frac{2}{3}[\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)]$


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