Telescoping Series Sum of 100 Terms

CAT 2017 Slot 2 · QA · Medium · Progressions

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If a1=12×5a_1 = \frac{1}{2 \times 5}, a2=15×8a_2 = \frac{1}{5 \times 8}, a3=18×11,…a_3 = \frac{1}{8 \times 11}, \dots, then a1+a2+a3+⋯+a100a_1 + a_2 + a_3 + \dots + a_{100} is

  1. A.

    25/151

  2. B.

    1/2

  3. C.

    1/4

  4. D.

    111/55

Answer

A

Explanation

The nn-th term an=1(3n−1)(3n+2)a_n = \frac{1}{(3n - 1)(3n + 2)}.

Using partial fractions: an=13(13n−1−13n+2)a_n = \frac{1}{3} \left( \frac{1}{3n - 1} - \frac{1}{3n + 2} \right)

Sum of first 100 terms: S100=13[(12−15)+(15−18)+⋯+(1299−1302)]S_{100} = \frac{1}{3} \left[ \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{8}\right) + \dots + \left(\frac{1}{299} - \frac{1}{302}\right) \right]

All middle terms cancel out: S100=13(12−1302)=13(151−1302)=13×150302=50302=25151S_{100} = \frac{1}{3} \left( \frac{1}{2} - \frac{1}{302} \right) = \frac{1}{3} \left( \frac{151 - 1}{302} \right) = \frac{1}{3} \times \frac{150}{302} = \frac{50}{302} = \frac{25}{151}.

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