Question

The value of $\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\dots .+\frac{1}{(4n-1)(4n+3)}$ is

• A. $\frac{n}{7(4n-1)}$
• B. $\frac{n}{3(4n+3)}$
• C. $\frac{1}{3(4n-1)}$
• D. $\frac{1}{3(4n-3)}$

Answer 1

Answer: (B)

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

$\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\dots .+\frac{1}{(4n-1)(4n+3)}$

$T_n=\frac{1}{(4n-1)(4n+3)}=\frac{1}{4}[\frac{1}{4n-1}-\frac{1}{4n+3}]$

$T_1=\frac{1}{4}[\frac{1}{3}-\frac{1}{7}]$

$T_2=\frac{1}{4}[\frac{1}{7}-\frac{1}{11}]$

$T_n=\frac{1}{4}[\frac{1}{4n-1}-\frac{1}{4n+3}]$

$Sn=\frac{1}{4}[\frac{1}{3}-\frac{1}{4n+3}]=\frac{n}{3(4n+3)}$