The sum of the infinite series 1+frac{5}{6}+f | vedclass

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

Question

(C) Let $S = 1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \ldots$ $\frac{1}{6}S = \frac{1}{6} + \frac{5}{6^{2}} + \frac{

Vedclass · Accepted Answer

$\frac{288}{125}$

Vedclass · Answer

(C) Let $S = 1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \ldots$ $\frac{1}{6}S = \frac{1}{6} + \frac{5}{6^{2}} + \frac{12}{6^{3}} + \frac{22}{6^{4}} + \ldots$ Subtracting the two equations: $\frac{5}{6}S = 1 + \frac{4}{6} + \frac{7}{6^{2}} + \frac{10}{6^{3}} + \frac{13}{6^{4}} + \ldots$ $\frac{5}{36}S = \frac{1}{6} + \frac{4}{6^{2}} + \frac{7}{6^{3}} + \frac{10}{6^{4}} + \ldots$ Subtracting again: $(\frac{5}{6} - \frac{5}{36})S = 1 + \frac{3}{6} + \frac{3}{6^{2}} + \frac{3}{6^{3}} + \ldots$ $\frac{25}{36}S = 1 + \frac{\frac{3}{6}}{1 - \frac{1}{6}} = 1 + \frac{3}{6} 	imes \frac{6}{5} = 1 + \frac{3}{5} = \frac{8}{5}$ $S = \frac{8}{5} 	imes \frac{36}{25} = \frac{288}{125}$

The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^{2}}+\frac{22}{6^{3}}+\frac{35}{6^{4}}+\frac{51}{6^{5}}+\frac{70}{6^{6}}+\ldots$ is equal to

Similar Questions

The sum of the first $n$ terms of the series $\frac{3}{5}+\frac{21}{25}+\frac{117}{125}+\ldots$ is

Let $\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty$ and $\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots \infty$. Then the value of $(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_{5}(\beta)}$ is equal to:

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

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + \dots$ is

Let $b_1, b_2, \dots, b_n$ be a geometric sequence such that $b_1 + b_2 = 1$ and $\sum\limits_{k = 1}^\infty b_k = 2$. Given that $b_2 < 0$,then the value of $b_1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module