Question

What is the sum of the following series:
$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + + \frac{1}{100 \times 101}$

• A. $\frac{99}{100}$
• B. $\frac{1}{100}$
• C. $\frac{100}{101}$
• D. $\frac{101}{102}$

Hint:

When you see terms like $1/(n(n+1))$, always try partial fractions and look for telescoping cancellation. It simplifies entire series neatly.

Answer 1

The Correct Option is C

Each term is of the form $\frac{1}{n(n+1)}$. We use partial fractions:
$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$
Thus, the entire sum becomes:
$\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + + \left( \frac{1}{100} - \frac{1}{101} \right)$
Now, this is a telescoping series — most terms cancel out:
$= \frac{1}{1} - \frac{1}{101} = 1 - \frac{1}{101} = \frac{100}{101}$
So the final answer is (C) $\frac{100}{101}$