The value of sumlimits_{n = 0}^infty {frac{{{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $S = \sum_{n=0}^{\infty} \frac{(n+1)^2}{7^n} = \frac{1^2}{7^0} + \frac{2^2}{7^1} + \frac{3^2}{7^2} + \frac{4^2}{7^3} + \dots$Multiply by $\fra

Vedclass · Accepted Answer

$\frac{49}{27}$

Vedclass · Answer

(A) Let $S = \sum_{n=0}^{\infty} \frac{(n+1)^2}{7^n} = \frac{1^2}{7^0} + \frac{2^2}{7^1} + \frac{3^2}{7^2} + \frac{4^2}{7^3} + \dots$Multiply by $\frac{1}{7}$:$\frac{S}{7} = \frac{1^2}{7^1} + \frac{2^2}{7^2} + \frac{3^2}{7^3} + \dots$Subtract the two equations:$S - \frac{S}{7} = 1 + \frac{2^2 - 1^2}{7^1} + \frac{3^2 - 2^2}{7^2} + \frac{4^2 - 3^2}{7^3} + \dots$$\frac{6S}{7} = 1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$This is an arithmetico-geometric series. Let $T = 1 + \frac{3}{7} + \frac{5}{7^2} + \frac{7}{7^3} + \dots$$\frac{T}{7} = \frac{1}{7} + \frac{3}{7^2} + \frac{5}{7^3} + \dots$Subtracting these:$T - \frac{T}{7} = 1 + \frac{2}{7} + \frac{2}{7^2} + \frac{2}{7^3} + \dots$$\frac{6T}{7} = 1 + \frac{2/7}{1 - 1/7} = 1 + \frac{2/7}{6/7} = 1 + \frac{1}{3} = \frac{4}{3}$$T = \frac{4}{3} 	imes \frac{7}{6} = \frac{14}{9}$Since $\frac{6S}{7} = T$,we have $\frac{6S}{7} = \frac{14}{9} \Rightarrow S = \frac{14}{9} 	imes \frac{7}{6} = \frac{98}{54} = \frac{49}{27}$.

The value of $\sum\limits_{n = 0}^\infty {\frac{{{{(n + 1)}^2}}}{{{7^n}}}}$ is -

Similar Questions

The sum of $(n - 1)$ terms of the series $1 + (1 + 3) + (1 + 3 + 5) + \dots$ is

$2 + 4 + 7 + 11 + 16 + \dots$ to $n$ terms =

$\sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1) =$

If $n$ is odd or even,the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ is given by:

Find the $9^{\text{th}}$ term in the sequence whose $n^{\text{th}}$ term is given by $a_{n} = (-1)^{n-1} n^{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module