Question

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

• A. $\frac{425}{216}$
• B. $\frac{429}{216}$
• C. $\frac{288}{125}$
• D. $\frac{280}{125}$

Answer 1

Answer: (C)

$\frac{288}{125}$

Answer 2

The correct answer is (C) : $\frac{288}{125}$
$S = 1 + \frac{5}{6} + \frac{12}{6^2} + \frac{22}{6^3} + \frac{35}{6^4} + .....$
$\frac{S}{6} = \frac{1}{6} + \frac{5}{6^2} + \frac{12}{6^3} + \frac{22}{6^4} + ......$

On subtraction
$\frac{5}{6} S = 1 + \frac{4}{6} + \frac{7}{6^2} + \frac{10}{6^3} + \frac{13}{6^4} + ......$
$\frac{5}{36} S = 1 + \frac{4}{6^2} + \frac{7}{6^3} + \frac{10}{6^4} + \frac{13}{6^5} + ......$

On Subtraction
$\frac{25}{36} S = 1 + \frac{3}{6} + \frac{3}{6^2} + \frac{3}{6^3} + ........ = \frac{8}{5}$
$S = \frac{288 }{125}$