The value of

Solution:

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

For $\sum_{r=1}^{n}(2 r-1)=1+3+5+7+\ldots \ldots+2 n+1$

Here  first term is 1 

last term is 2n − 1

i.e sum is $\frac{n}{2}$ (first term + last term)

i.e $\frac{n}{2}(1+2 n-1)$

$=\frac{n}{2}(2 n)$

i. e $\sum_{r=1}^{n}(2 r-1)=n^{2} \quad \ldots(1)$

also $\sum_{r=1}^{n} \frac{1}{2^{r}}=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots+\frac{1}{2^{n}}$

first term $a=\frac{1}{2}, r=$ common ratio is $\frac{1}{2}$

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

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

$\sum_{r=1}^{n} \frac{1}{2^{r}}=S \mathrm{um}=1-\frac{1}{2^{n}}$

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

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