Find the sum of n terms of the series whose

We need to find the sum of $n$ terms of series whose $r^{\text {th }}$ term is $r+2^{r}$.

$a_{r}=r+2^{r}$

So, $n^{\text {th }}$ term, $a_{n}=n+2^{n}$

So, we can find the sum of the series by using summation of the $n^{\text {th }}$ term of the given series.

$S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k+2^{k}$

$S_{n}=\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k+\sum_{k=1}^{n} 2^{k} \rightarrow$ (1)

Note:

I. Sum of first n natural numbers, 1 + 2 +3+…n

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

Second term in $(2)$ is a GP, with first term a $=2$, common ratio $r=2$.

Sum of $n$ terms of GP, with the first term, $a$, common ratio, $r$,

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

So, the sum of given GP, with a = 2, r = 2

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

$S_{n}=\sum_{k=1}^{n} k+\sum_{k=1}^{n} 2^{k}$

The required sum,

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

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

The sum of $n$ terms of the series whose $r^{\text {th }}$ term is $(r+2 r)$,

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