Sum of Infinite Series

CAT 2024 Slot 2 · QA · Medium · Algebra

The sum of the infinite series 15(1517)+(15)2((15)2(17)2)+(15)3((15)3(17)3)+\frac{1}{5}\left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left(\left(\frac{1}{5}\right)^2 - \left(\frac{1}{7}\right)^2\right) + \left(\frac{1}{5}\right)^3\left(\left(\frac{1}{5}\right)^3 - \left(\frac{1}{7}\right)^3\right) + \cdots is equal to

  1. A.

    \frac{5}{408}

  2. B.

    \frac{7}{816}

  3. C.

    \frac{7}{408}

  4. D.

    \frac{5}{816}

Answer

A

Explanation

The given series can be rewritten as: n=1(15)n[(15)n(17)n]=n=1[(125)n(135)n]\sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n \left[\left(\frac{1}{5}\right)^n - \left(\frac{1}{7}\right)^n\right] = \sum_{n=1}^{\infty} \left[\left(\frac{1}{25}\right)^n - \left(\frac{1}{35}\right)^n\right]

This splits into two infinite geometric series: S1=n=1(125)n=1/2511/25=124S_1 = \sum_{n=1}^{\infty} \left(\frac{1}{25}\right)^n = \frac{1/25}{1 - 1/25} = \frac{1}{24} S2=n=1(135)n=1/3511/35=134S_2 = \sum_{n=1}^{\infty} \left(\frac{1}{35}\right)^n = \frac{1/35}{1 - 1/35} = \frac{1}{34}

Sum =S1S2=124134=3424816=10816=5408= S_1 - S_2 = \frac{1}{24} - \frac{1}{34} = \frac{34 - 24}{816} = \frac{10}{816} = \frac{5}{408}.

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