Question

The sum to $n$ terms of the series $1+2(1+\frac{1}{n})+3{(1+\frac{1}{n})}^2+...$ is given by

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

Answer 1

The correct option is A $n^2$

We have

$Sm=1+2(1+1/n)+3(1+1/n{)}^2+4(1+1/n{)}^3+...\dots ..+m(1+1/n{)}^{m-1}$ …………(A)

multiplying each term by $(1+1/n)$ we get,

$Sm(1+1/n)=1(1+1/n)+2(1+1/n{)}^2+3(1+1/n{)}^3+4(1+1/n{)}^4+...\dots \dots +(m-1\left)\right(1+1/n{)}^{m-1}+m(1+1/n{)}^m$ ……………(B)

Subtracting (B) from (A), we get

$-\frac{Sm}{n}=1+(1+\frac{1}{n})+(1+1/n{)}^2+(1+1/n{)}^3+...\dots ..+(1+1/n{)}^{m-1}-m(1+1/n{)}^m$

$=\frac{(1+1/n{)}^m-1}{1+\frac{1}{n}-1}-m(1+1/n{)}^m$

(or)

$Sm=mn(1+1/n{)}^m-n^2((1+1/n{)}^m-1)$

and if $m=n$

$Sn=n^2(1+1/n{)}^n-n^2((1+1/n{)}^n-1)$

$Sn=n^2$

$\therefore$ So, the answer is A. $n^2$.