Question

If $S$ is sum of $\frac{5}{1^2 \cdot 2^2 \cdot 3}+\frac{8}{2^2 \cdot 3^2 \cdot 4}+\frac{11}{3^2 \cdot 4^2 \cdot 5}+\frac{14}{4^2 \cdot 5^2 \cdot 6}+\ldots \ldots \infty$ terms, then $S$ is equal to

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

Answer: (B)

$\Theta \quad T_n=\frac{(3 n+2)}{n^2(n+1)^2(n+2)}=\frac{\left(n^2+3 n+2\right)-n^2}{n^2(n+1)^2(n+2)}$
$=\frac{1}{ n ^2( n +1)}-\frac{1}{( n +1)^2( n +2)} $
$\therefore \quad S =\left(\frac{1}{1^2 \cdot 2}-\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3}-\frac{1}{3^2 \cdot 4}\right)+\left(\frac{1}{3^2 \cdot 4}-\frac{1}{4^2 \cdot 5}\right) \ldots \ldots \infty \text { terms } $
$=\frac{1}{2}$