Question

Sum to $n$ terms of the series $\frac{1}{\left(1.3\right)-\frac{3}{8}}+\frac{1}{\left(2.5\right)-\frac{3}{8}}+\frac{1}{\left(3.7\right)-\frac{3}{8}}+\cdots$ is

• A. $\frac{4n}{3(4n+3)}$
• B. $\frac{2n}{4n+3}$
• C. $\frac{8n}{3(4n+3)}$
• D. $\frac{2n}{3(4n+3)}$

Answer 1

Answer: (C)

$\frac{8n}{3(4n+3)}$

Let, $S_n=\frac{1}{\left(1.3\right)-3/8}+\frac{1}{\left(2.5\right)-3/8}+\frac{1}{\left(3.7\right)-3/8}+\cdots$
Now, $a_n=\frac{1}{n(2n+1)-3/8}=\frac{8}{16n^2+8n-3}=\frac{8}{(4n-1)(4n+3)}$

$\Rightarrow a_n=2[\frac{1}{(4n-1)}-\frac{1}{(4n+3)}]$
$S_n={\sum }_{n=1}^n{a}_n=2{\sum }_{n=1}^n[\frac{1}{(4n-1)}-\frac{1}{(4n+3)}]$
$\Rightarrow S_n=2[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....-\frac{1}{(4n+3)}]=2[\frac{1}{3}-\frac{1}{(4n+3)}]$
$\Rightarrow S_n=\frac{8n}{3(4n+3)}$
Ans: C