Question

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

Answer 1

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

We have:
$$\begin{gathered} \sum _{r=1}^n\left\{(2r-1)\mathit{}a+\frac{1}{b^r}\right\} \\ =\sum _{r=1}^n\left\{2ra-a+\frac{1}{b^r}\right\} \\ =\sum _{r=1}^{n}2ar-\sum _{r=1}^{n}a+\sum _{r=1}^n\frac{1}{b^r} \\ =an\left(n+1\right)-an+\frac{\left(1-b^n\right)}{\left(1-b\right)b^n} \\ =a{n}^2+\frac{\left(b^n-1\right)}{\left(b-1\right)b^n} \end{gathered}$$