What is the sum of the series 1 + frac{2}{3} | vedclass

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

Question

(D) Let $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots \infty \dots (1)$Multiplying by $\frac{1}{3}$,we get $\frac{1}{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots \infty \dots (1)$Multiplying by $\frac{1}{3}$,we get $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots \infty \dots (2)$Subtracting $(2)$ from $(1)$:$S - \frac{1}{3}S = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{3^2} - \frac{2}{3^2}) + (\frac{10}{3^3} - \frac{6}{3^3}) + \dots$$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots$$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{3^2} (1 + \frac{1}{3} + \frac{1}{3^2} + \dots)$Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$:$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{4}{9} (\frac{1}{1 - 1/3}) = 1 + \frac{1}{3} + \frac{4}{9} (\frac{3}{2}) = 1 + \frac{1}{3} + \frac{2}{3} = 2$$\frac{2}{3}S = 2 \implies S = 3$.

What is the sum of the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$?

Similar Questions

The sum of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ up to $20$ terms is

If $n$ is the smallest natural number such that $n+2n+3n+\ldots+99n$ is a perfect square,then the number of digits of $n^2$ is

Find the sum of the series $5^{2} + 6^{2} + 7^{2} + \ldots + 20^{2}$.

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

Find the sum to $n$ terms of the series: $5+11+19+29+41 + \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module