Question

The value of $\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+....+\frac{1}{99\times 100}$ is ____.

• A. Less than $\frac{99}{100}$
• B. Equal to $\frac{99}{100}$
• C. Greater than $\frac{100}{99}$
• D. Equal to $\frac{100}{99}$

Answer 1

Answer: (B)

Equal to $\frac{99}{100}$

$nth$ term of the series can be written as following,

$T_n=\frac{1}{n\times (n+1)}=\frac{(n+1)-n}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$

If $S_n$ is sum of the first n terms of the given series,

Then, $S_{99}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots -\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}$