Question

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

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

$Wehave,{\sum }_{r=1}^n\left\{\right(2r-1)a+\frac{1}{b^r}\}={\sum }_{r=1}^n\{2ra-a+\frac{1}{b^r}\}={\sum }_{r=1}^{n}2ra-{\sum }_{r=1}^{n}a+{\sum }_{r=1}^n\frac{1}{b^r}=an(n+1)-an+\frac{(1-b^n)}{(1-b)b^n}=a{n}^2+\frac{(b^n-1)}{(b-1)b^n}$