Sequences And Series — Easy JEE math MCQ
The sum of the series \(\sum_{n=1}^{99} \frac{1}{n(n+1)}\) is:
- A. \(\frac{99}{100}\)
- B. \(\frac{98}{99}\)
- C. \(\frac{100}{101}\)
- D. \(\frac{97}{98}\)
Solution
Using partial fractions:
\[
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
\]
Therefore:
\[
\sum_{n=1}^{99} \frac{1}{n(n+1)} = \sum_{n=1}^{99} \left(\frac{1}{n} - \frac{1}{n+1}\right)
\]
This is a telescoping series:
\[
= \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \ldots + \left(\frac{1}{99} - \frac{1}{100}\right)
\]
\[
= 1 - \frac{1}{100} = \frac{99}{100}
\]
So the correct option is A.