Question

The value of $\frac{1}{(1+\mathrm{a})(2+\mathrm{a})}+\frac{1}{(2+\mathrm{a})(3+\mathrm{a})}+\frac{1}{(3+\mathrm{a})(4+\mathrm{a})}+.....\infty$is (where $\mathrm{a}$ is a constant)

• A. $\frac{1}{1+a}$
• B. $\frac{1}{2+a}$
• C. $\infty$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{1+a}$

The explanation of the correct answer.

Find the value of the given expression.

Given: $\frac{1}{(1+\mathrm{a})(2+\mathrm{a})}+\frac{1}{(2+\mathrm{a})(3+\mathrm{a})}+\frac{1}{(3+\mathrm{a})(4+\mathrm{a})}+.....\infty$

Rearrange the term like this

$\frac{1}{\left(1+\mathrm{a}\right)}-\frac{1}{\left(2+\mathrm{a}\right)}=\frac{2+\mathrm{a}-1-\mathrm{a}}{\left(1+\mathrm{a}\right)\left(1+2\mathrm{a}\right)}=\frac{1}{\left(1+\mathrm{a}\right)\left(1+2\mathrm{a}\right)}$

Hence the expression change into

$$\begin{gathered} \Rightarrow \frac{1}{\left(1+\mathrm{a}\right)}-\frac{1}{\left(2+\mathrm{a}\right)}+\frac{1}{\left(2+\mathrm{a}\right)}-\frac{1}{\left(3+\mathrm{a}\right)}+\frac{1}{\left(3+\mathrm{a}\right)}-\frac{1}{\left(4+\mathrm{a}\right)}...... \\ =\frac{1}{\left(1+\mathrm{a}\right)}\left[\mathrm{Cancel}\mathrm{all}\mathrm{like}\mathrm{terms}\right] \end{gathered}$$

Hence $o\mathrm{ption}\left(\mathrm{A}\right)$ is the correct answer.