Sum of Fractional Series

CAT 2000 Slot 1 · QA · Medium · Algebra

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What is the value of the following expression?

(122−1)+(142−1)+(162−1)+⋯+(1202−1)\left(\frac{1}{2^2 - 1}\right) + \left(\frac{1}{4^2 - 1}\right) + \left(\frac{1}{6^2 - 1}\right) + \dots + \left(\frac{1}{20^2 - 1}\right)

  1. A.

    919\frac{9}{19}

  2. B.

    1019\frac{10}{19}

  3. C.

    1021\frac{10}{21}

  4. D.

    1121\frac{11}{21}

Answer

C

Explanation

The general term is Tk=1(2k)2−1=1(2k−1)(2k+1)=12(12k−1−12k+1)T_k = \frac{1}{(2k)^2 - 1} = \frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right). Summing from k=1k=1 to 1010 gives: S=12(1−121)=12×2021=1021S = \frac{1}{2}\left(1 - \frac{1}{21}\right) = \frac{1}{2} \times \frac{20}{21} = \frac{10}{21}.

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