Question

If $\sum _{r=1}^n{t}_r=\frac{n(n+1)(n+2)(n+3)}{8}$, then $\sum _{r=1}^n\frac{1}{t_r}$ equals

• A. $-(\frac{1}{(n+1)(n+2)}-\frac{1}{2})$
• B. $(\frac{1}{(n+1)(n+2)}-\frac{1}{2})$
• C. $(\frac{1}{(n+1)(n+2)}+\frac{1}{2})$
• D. $(\frac{1}{(n-1)(n-2)}+\frac{1}{2})$.

Answer 1

Answer: (A)

$-(\frac{1}{(n+1)(n+2)}-\frac{1}{2})$

Let $S_n={\sum }_{r=1}^n{t}_r$

ie, $t_n=S_n-S_{n-1}=\frac{n(n+1)(n+2)(n+3)}{8}-\frac{(n-1)n(n+1)(n+2)}{8}=\frac{n(n+1)(n+2)(n+3-(n-1\left)\right)}{8}=\frac{n(n+1)(n+2)}{2}$

Let, $\frac{1}{t_n}=\frac{2}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}$

$\Rightarrow \frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}=\frac{A(n+1)+Bn}{n(n+1)}+\frac{C}{n+2}=\frac{\left(\right(A+B)n+A)(n+2)+Cn(n+1)}{n(n+1)(n+2)}$

$\Rightarrow \frac{(A+B+C)n^2+(3A+2B+C)n+2A}{n(n+1)(n+2)}=\frac{2}{n(n+1)(n+2)}$

ie,$A=1,B=(-2),C=1$

$\therefore {\sum }_{r=1}^n\frac{1}{t_r}={\sum }_{r=1}^n(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2})=(\frac{1}{1}+-\frac{2}{2}+\frac{1}{2}++\frac{1}{3}-\frac{2}{3}++\frac{1}{3}++\frac{1}{4}-\frac{2}{4}+\frac{1}{4}+...++\frac{1}{n}-\frac{2}{n}+\frac{1}{n}++\frac{1}{n+1}-\frac{2}{n+1}+\frac{1}{n+2})$

$\Rightarrow {\sum }_{r=1}^n\frac{1}{t_r}=\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}=\frac{1}{2}-\frac{1}{(n+1)(n+2)}$

So, Option A is correct