# The value of

Question:

The value of $\sum_{r=1}^{n}\left\{(2 r-1) a+\frac{1}{b^{r}}\right\}$ is equal to

(a) $a n^{2}+\frac{b^{n-1}-1}{b^{n-1}(b-1)}$

(b) $a n^{2}+\frac{b^{n}-1}{b^{n}(b-1)}$

(c) $a n^{3}+\frac{b^{n-1}-1}{b^{n}(b-1)}$

(d) none of these

Solution:

(b) $a n^{2}+\frac{b^{n}-1}{b^{n}(b-1)}$

we have,

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

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

$=\sum_{r=1}^{n} 2 a r-\sum_{r=1}^{n} a+\sum_{r=1}^{n} \frac{1}{b^{r}}$

$=a n(n+1)-a n+\frac{\left(1-b^{n}\right)}{(1-b) b^{n}}$

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