Question

Sum to n terms of the series $\frac{1}{5!}+\frac{1!}{6!}+\frac{2!}{7!}+\frac{3!}{8!}+.....$ is

• A. $\frac{2}{5!}-\frac{1}{(n+1)!}$
• B. $\frac{1}{4}(\frac{1}{4!}-\frac{n!}{(n+4)!})$
• C. $\frac{1}{4}(\frac{1}{3!}-\frac{3!}{(n+2)!})$
• D. $\frac{1}{4}(\frac{1}{4!}+\frac{n!}{(n+4)!})$

Answer 1

The correct option is B $\frac{1}{4}(\frac{1}{4!}-\frac{n!}{(n+4)!})$

We have $t_r=\frac{(r-1)!}{(r+4)!}$ and $t_{r+1}=\frac{r!}{(r+5)!}$

Now, $r{t}_r-(r+5)t_{r+1}=\frac{r!}{(r+4)!}-\frac{r!}{(r+4)!}=0$

$\Rightarrow r{t}_r-(r+1)t_{r+1}=4t_{r+1}\Rightarrow 4\sum _{r=1}^{n-1}{t}_{r+1}=\sum _{r=1}^{n-1}[r{t}_r-(r+1\left)t_{r+1}\right]$

$\Rightarrow 4(t_2+t_3+....+t_n)=1t_1-n{t}_n$

$\Rightarrow 4(t_1+t_2+.....t_n)=5t_1-n{t}_n$$=5\left(\frac{0!}{5!}\right)-\frac{n(n-1)!}{(n+4)!}$

$\Rightarrow t_1+t_2+....t_n=\frac{1}{4}[\frac{1}{4!}-\frac{n!}{(n+4)!}]$

Hence, option 'B' is correct.