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

Solution:

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

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

Now, we need to find the sum of this series, $\mathrm{Sn}$.

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

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

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

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

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

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

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,

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

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

$=\left(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)-3 \mathrm{n}(\mathrm{n}+1)+4 \mathrm{n}}{2}\right)$

On simplifying,

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

Hence, the sum of the series, $\mathrm{S}_{\mathrm{n}}=\mathrm{n}\left(\mathrm{n}^{2}+1\right)$