Question

The value of ${\sum }_{r=1}^n\{(2r-1)a+\frac{1}{b^r}\}$ 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

Answer 1

The correct option is B $a{n}^2+\frac{b^n-1}{b^n(b-1)}$

Let $S={\sum }_{r=1}^n\{(2r-1)a+\frac{1}{b^r}\}$
$\Rightarrow S=2a{\sum }_{r=1}^{n}r-a{\sum }_{r=1}^n+{\sum }_{r=1}^n\left(\frac{1}{b^r}\right)$
$\Rightarrow S=an(n+1)-an+\frac{1}{b}\left[\frac{{\left(1/b\right)}^n-1}{\left(1/b\right)-1}\right]$
$\Rightarrow S=a{n}^2+\frac{b^n-1}{b^n(b-1)}$
Ans: B