Sum of Infinite Series

CAT 2024 Slot 2 · Quantitative Ability · Medium · Algebra

The sum of the infinite series 15(15−17)+(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=1/251−1/25=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=1/351−1/35=134S_2 = \sum_{n=1}^{\infty} \left(\frac{1}{35}\right)^n = \frac{1/35}{1 - 1/35} = \frac{1}{34}

Sum =S1−S2=124−134=34−24816=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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