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UPSC CDS 1 2021 Math Paper

Question : 80 of 100

Marks: +1, -0

\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdot \cdot \cdot +\frac{1}{n(n+1)}=\frac{99}{100}

then what is the value of n?

Solution:

Concept:

This series uses telescoping cancellation where each term splits into a difference of fractions.

Explanation:

Rewrite each term:

\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}

The sum from

k=1

n

becomes:

\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)

All intermediate terms cancel, leaving

1 - \frac{1}{n+1}

Set this equal to

\frac{99}{100}

1 - \frac{1}{n+1} = \frac{99}{100}

Rearrange:

\frac{1}{n+1} = 1 - \frac{99}{100} = \frac{1}{100}

Take reciprocals:

n+1 = 100

, so

n = 99

Answer:

The value of

n

99

, which corresponds to option B.

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