Question

Let the $n$ terms of series is $\frac{1}{\mathrm{1.2.3.4}}+\frac{1}{\mathrm{2.3.4.5}}+\frac{1}{\mathrm{3.4.5.6}}+\dots \dots$ then

• A. $T_n=\frac{1}{n(n+1)(n+2)(n+3)}$
• B. $S_{\infty }=\frac{1}{6}$
• C. $S_{\infty }=\frac{1}{18}$
• D. $S_n=\frac{1}{3(n+1)(n+2)(n+3)}$

Answer 1

Answer: (A), (C)

$S_{\infty }=\frac{1}{18}$

General term of the series is
$T_n=\frac{1}{n(n+1)(n+2)(n+3)}$
$T_n=\frac{1}{3}(\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)})$
Let $V_{n-1}=\frac{1}{n(n+1)(n+2)}$
$\therefore T_n=\frac{1}{3}[V_{n-1}-V_n]$
Hence
$S_n=\frac{1}{3}[V_0-V_n]=\frac{1}{3}[\frac{1}{\mathrm{1.2.3}}-\frac{1}{(n+1)(n+2)(n+3)}]$
$S_n=\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}{S}_{\infty }=\frac{1}{18}$