The sum to n terms of the series 1+2(1+(1/n))+3(1+(1/n))2+ ldots is given by

Question

The sum to n terms of the series 1+2(1+(1/n))+3(1+(1/n))2+ ldots is given by (A) n2 (B) n(n+1) (C) n(1+1 / n)2 (D) None of these

Tardigrade Admin · Accepted Answer

Let S be the sum of n terms of the given series and x=1+1 / n. Then, S=1+2 x+3 x2+4 x3+ ldots+ n xn-1 ⇒ x S=x+2 x2+3 x3+ ldots+(n-1) xn-1+n xn ∴ S-x S=1+[x+x2+ ldots+xn-1]-n xn ⇒ S(1-x)=(1-xn/1-x)-n xn ⇒ S(-1 / n)=-n(1-(1+1 / n)n)-n(1+1 / n)n ⇒ (1/n) S=n[1-(1+1 / n)n+(1+1 / n)n]=n ⇒ S=n2

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

$n^{2}$

$n (n + 1)$

$n (1 + 1/ n)^{2}$

None of these

Q. The sum to n terms of the series 1+2(1+n1​)+3(1+n1​)2+… is given by

n2

n(n+1)

n(1+1/n)2

None of these

Let S be the sum of n terms of the given series and x=1+1/n. Then, S=1+2x+3x2+4x3+…+nxn−1 ⇒xS=x+2x2+3x3+…+(n−1)xn−1+nxn ∴S−xS=1+[x+x2+…+xn−1]−nxn ⇒S(1−x)=1−x1−xn​−nxn ⇒S(−1/n)=−n(1−(1+1/n)n)−n(1+1/n)n ⇒n1​S=n[1−(1+1/n)n+(1+1/n)n]=n ⇒S=n2

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

$n^{2}$

$n (n + 1)$

$n (1 + 1/ n)^{2}$