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

It is given in the question that the nth term of the series,

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

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

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

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

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

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

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

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,

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

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

The second term in the equation $\sum_{n=1}^{n}\left(3^{n}\right)=3^{1}+3^{2}+\cdots 3^{n}$

Forms a GP, with the common ratio, r = 3.

Sum of $n$ terms of a GP, $a, a r, a r^{2}, a r^{3} \ldots a r^{n} .$

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}$

Here, a= 3, r = 3;

So,

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

Substitute (2) in (1);

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

Hence, the sum of the series, $S_{n}=\left(\frac{n(n+1)}{2}\right)^{2}+\frac{3\left(3^{n}-1\right)}{2}$