# Find the sum of the series whose nth term is given by:

Question:

Find the sum of the series whose nth term is given by:

$\left(2 n^{2}-3 n+5\right)$

Solution:

It is given in the question that the $n^{\text {th }}$ term of the series,

$a_{n}=2 n^{2}-3 n+5$

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

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

$\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(2 \mathrm{n}^{2}-3 \mathrm{n}+5\right)$

$=\sum_{\mathrm{n}=1}^{\mathrm{n}}\left(2 \mathrm{n}^{2}\right)-\sum_{\mathrm{n}=1}^{\mathrm{n}}(3 \mathrm{n})+\sum_{\mathrm{n}=1}^{\mathrm{n}}(5)$

$=2 \sum_{\mathrm{n}=1}^{\mathrm{n}}\left(\mathrm{n}^{2}\right)-3 \sum_{\mathrm{n}=1}^{\mathrm{n}}(\mathrm{n})+\sum_{\mathrm{n}=1}^{\mathrm{n}}(5)$

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

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

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,

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

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

$S_{n}=2 \sum_{n=1}^{n}\left(n^{2}\right)-3 \sum_{n=1}^{n}(n)+\sum_{n=1}^{n}(5)$

$\mathrm{S}_{\mathrm{n}}=2\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\right)-3\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)+5 \mathrm{n}$

$=\left(\frac{2 n(n+1)(2 n+1)-9 n(n+1)+30 n}{6}\right)$

$=\left(\frac{4 n^{3}-3 n^{2}+23 n}{6}\right)$

$=\frac{n}{6}\left(4 n^{2}-3 n+23\right)$

Hence, the sum of the series, $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{6}\left(4 \mathrm{n}^{2}-3 \mathrm{n}+23\right)$