Question

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\dots .+\frac{1}{n(n+1)}equals$

• A. $\frac{1}{n(n+1)}$
• B. $\frac{n}{(n+1)}$
• C. $\frac{2n}{(n+1)}$
• D. $\frac{2}{n(n+1)}$

Answer 1

The correct option is B

$\frac{n}{(n+1)}$

Explanation for the correct option:

Rearrange the given expression as sum of difference of two terms

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\dots .+\frac{1}{n(n+1)}$

$=\frac{(2-1)}{1.2}+\frac{(3-2)}{2.3}+\frac{(4-3)}{3.4}+\dots .+\frac{(n+1)-n}{n(n+1)}$

$=\frac{2}{1.2}–\frac{1}{1.2}+\frac{3}{2.3}–\frac{2}{2.3}+\frac{4}{3.4}–\frac{3}{3.4}+.....+\frac{(n+1)}{n(n+1)}–\frac{n}{n(n+1)}$

$=1–\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}–\frac{1}{4}+\dots .+\frac{1}{n}–\frac{1}{(n+1)}$

$=1–\frac{1}{(n+1)}$ [ Rest all terms got cancelled ]

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

$=\frac{\mathbf{n}}{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}$

Hence, Option ‘B’ is Correct