The sum of the infinite series 1+(5)/(6)+(12)/(6^2)+(22)/(6^3)+(35)/(6^4)+(51)/(6^5)+(70)/(6^6)+ . . is equal to:

Correct answer: C. (288)/(125) Let S=1+(5)/(6)+(12)/(6^2)+(22)/(6^3)+(35)/(6^4)+ .....(i) (S)/(6)=(1)/(6)+(5)/(6^2)+(12)/(6^3)+(22)/(6^4)+ ....(ii) On subtracting (i) from (ii) (5)/(6) S=1+(4)/(6)+(7)/(6^2)+(10)/(6^3)+(13)/(6^4)+ (5)/(36) S=(1)/(6)+(4)/(6^2)+(7)/(6^3)+(10)/(6^4)+(13)/(6^5)+ on subtraction (25)/(36) S=1+(3)/(6)+(3)/(6^2)+(3)/(6^3)+ =(8)/(5)S=(8)/(5) × (36)/(25)=(288)/(125)

BITSAT 2024Paper 2

Exam: BITSAT
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Sequences and Series

BITSAT 2024 · Paper 2 · Mathematics · Sequences and Series

The sum of the infinite series...

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:

Options

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

Correct answer

C. $\frac{288}{125}$

Step-by-step solution

Let

S=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\ldots

.....(i)

\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\ldots

....(ii) On subtracting (i) from (ii)

\frac{5}{6} S=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\cdots\frac{5}{36} S=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\frac{13}{6^5}+\cdots

on subtraction

\frac{25}{36} S=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\ldots=\frac{8}{5}S=\frac{8}{5} \times \frac{36}{25}=\frac{288}{125}

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