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Sum of Infinite Series

CAT 2023 Slot 3 · Quantitative Ability · Medium · Algebra

This is a medium Quantitative Ability question from the CAT 2023 Slot 3 paper. It tests Algebra. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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The value of 1+(1+13)14+(1+13+19)116+(1+13+19+127)164+⋯1 + \left(1 + \frac{1}{3}\right)\frac{1}{4} + \left(1 + \frac{1}{3} + \frac{1}{9}\right)\frac{1}{16} + \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\right)\frac{1}{64} + \cdots, is

  1. A.

    15/3

  2. B.

    16/11

  3. C.

    27/12

  4. D.

    15/8

Answer

B

Explanation

The nn-th term is Tn=(∑k=0n−113k)14n−1T_n = \left(\sum_{k=0}^{n-1} \frac{1}{3^k}\right) \frac{1}{4^{n-1}}.

Using GP sum formula: ∑k=0n−113k=1−(1/3)n1−1/3=32(1−13n)\sum_{k=0}^{n-1} \frac{1}{3^k} = \frac{1 - (1/3)^n}{1 - 1/3} = \frac{3}{2}\left(1 - \frac{1}{3^n}\right) Tn=32(1−13n)14n−1=32(14n−1−13⋅12n−1)T_n = \frac{3}{2}\left(1 - \frac{1}{3^n}\right)\frac{1}{4^{n-1}} = \frac{3}{2}\left(\frac{1}{4^{n-1}} - \frac{1}{3 \cdot 12^{n-1}}\right)

Sum of infinite series: S=∑n=1∞Tn=32(∑n=1∞14n−1−13∑n=1∞112n−1)S = \sum_{n=1}^\infty T_n = \frac{3}{2} \left( \sum_{n=1}^\infty \frac{1}{4^{n-1}} - \frac{1}{3} \sum_{n=1}^\infty \frac{1}{12^{n-1}} \right) S=32(11−1/4−13⋅11−1/12)=32(43−411)S = \frac{3}{2} \left( \frac{1}{1 - 1/4} - \frac{1}{3} \cdot \frac{1}{1 - 1/12} \right) = \frac{3}{2} \left( \frac{4}{3} - \frac{4}{11} \right) S=32⋅4⋅(13−111)=6×833=1611S = \frac{3}{2} \cdot 4 \cdot \left(\frac{1}{3} - \frac{1}{11}\right) = 6 \times \frac{8}{33} = \frac{16}{11}

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