Question

For first $n$ natural numbers we have the following results with usual notations $\sum _{r=1}^{n}r=\frac{n(n+1)}{2},\sum _{r=1}^n{r}^2=\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^n{r}^3={\left(\sum _{r=1}^{n}r\right)}^2$ If $a_1{a}_2....a_n\in A.P$ then sum to $n$ terms of the sequence $\frac{1}{a_1{a}_2},\frac{1}{a_2{a}_3},...\frac{1}{a_{n-1}{a}_n}$ is equal to $\frac{n-1}{a_1{a}_n}$
and the sum to $n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by $S_n=\frac{lr-a}{r-1}$ for $r\ne 1$ for $r=1$ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
$\frac{lr-a}{r-1}$ when $n\to \infty ,\left|r\right|<l$ where $l$ is the last term of $G.P.$ On the basis of above data answer the following questionsThe sum of the series $\frac{1}{2.3}+\frac{1}{3.4}+...$ to $n$ terms is?

• A. $\frac{n}{2(n+2)}$
• B. $\frac{n}{n+2}$
• C. $\frac{n+1}{2(n+2)}$
• D. None of these

Answer 1

The correct option is A $\frac{n}{2(n+2)}$

Let, $S_n=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{(n+1)(n+2)}$
Now, $t_r=\frac{1}{(r+1)(r+2)}=\frac{1}{r+1}-\frac{1}{r+2}$
$\therefore S_n={\sum }_{r=1}^n{t}_r={\sum }_{r=1}^n(\frac{1}{r+1}-\frac{1}{r+2})$
$\Rightarrow S_n=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{n+1}-\frac{1}{n+2})$
$\Rightarrow S_n=\frac{1}{2}-\frac{1}{n+2}=\frac{n}{2(n+2)}$
Ans: A