Question

Ths sum to infinity of the series, $1+2(1-\frac{1}{n})+3{(1-\frac{1}{n})}^2+...$ is

• A. $n^2$
• B. $n(n+1)$
• C. $n{(1+\frac{1}{n})}^2$
• D. None of these

Answer 1

The correct option is A $n^2$

Let $S=1+2(1-\frac{1}{n})+3({1-\frac{1}{n})}^2+...$ $...\left(1\right)$
$\therefore (1-\frac{1}{n})S=(1-\frac{1}{n})+2({1-\frac{1}{n})}^2+...$ $...\left(2\right)$
Now, subtracting $\left(1\right)$ by $\left(2\right)$; we get
$S-(1-\frac{1}{n})S=1+2(1-\frac{1}{n})+3({1-\frac{1}{n})}^2+...-(1-\frac{1}{n})-2({1-\frac{1}{n})}^2-...\infty$
$\Rightarrow S-\frac{(n-1)}{n}S=1+(1-\frac{1}{n})+{(1-\frac{1}{n})}^2+...\infty$
$\Rightarrow \frac{Sn-Sn+S}{n}=\frac{1}{1-(1-\frac{1}{n})}$
$\Rightarrow \frac{S}{n}=\frac{1}{\frac{1}{n}}$
$\Rightarrow S=n^2$