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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2023 · Paper 2 · Mathematics · Sequences and Series

The sum to infinite term of the series...

The sum to infinite term of the series 1+23+632+1033+1434+⋯1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\cdots isThe sum to infinite term of the series 1+(2)/(3)+(6)/(3^2)+(10)/(3^3)+(14)/(3^4)+ isThe sum to infinite term of the series 1+(2)/(3)+(6)/(3^2)+(10)/(3^3)+(14)/(3^4)+ is

Options

  1. A
    3
  2. B
    4
  3. C
    6
  4. D
    2

Correct answer

A.
3

Step-by-step solution

We have
The sum to infinite term of the series 1+(2)/(3)+(6)/(3^2)+(10)/(3^3)+(14)/(3^4)+ is

Multiplying both sides by 13\frac{1}{3}, we getThe sum to infinite term of the series 1+(2)/(3)+(6)/(3^2)+(10)/(3^3)+(14)/(3^4)+ is

Subtracting eqn. (ii) from eqn. (i), we get
23S=1+13+432+433+434+………∞⇒23S=43+432+433+434+………∞⇒23S=431−13=43×32⇒S=3\begin{aligned} & \frac{2}{3} S=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \ldots \infty \\ & \Rightarrow \frac{2}{3} S=\frac{\frac{4}{3}}{1-\frac{1}{3}}=\frac{4}{3} \times \frac{3}{2} \Rightarrow S=3\end{aligned}
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