Question

If constant term in the cubic expression $\sum _{k=1}^n[x-\frac{1}{k}][x-\frac{1}{k+1}][x-\frac{1}{k+2}]$ is $S_n$ , then $\underset{n\to \infty }{lim}{S}_n$ is

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

Answer 1

The correct option is D $\frac{-1}{4}$

Constant term of the given expression
$S_n=-\sum _{k=1}^n\frac{1}{k(k+1)(k+2)}$ (obtained by putting x = 0)
$S_n=-\frac{1}{2}\sum _{k=1}^n[\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}]$
$S_n=-\frac{1}{2}\sum _{k=1}^n(v_k-v_{k+1})$. $(let{v}_k=\frac{1}{k(k+1)})$
$S_n=-\frac{1}{2}\left[\right(v_1-v_2)+(v_2-v_3)+.........+(v_n-v_{n+1}\left)\right]$
$S_n=-\frac{1}{2}[\frac{1}{1.2}-\frac{1}{(n+1)(n+2)}]$
As $n\to \infty ,S_{\infty }=-\frac{1}{4}$