Question

If in a series $t_n=\frac{n}{(n+1)!}$, then ${\sum }_{n=1}^{10}{t}_n$ is equal to

• A. $\frac{20!-1}{20!}$
• B. $\frac{11!-1}{11!}$
• C. $\frac{1}{2(n-1)!}$
• D. $1+\frac{1}{21!}$

Answer 1

Answer: (B)

$\frac{11!-1}{11!}$

Given:

$t_n=\frac{n}{(n+1)!}$

$t_n=\frac{n+1-1}{(n+1)!}$
Splitting the fraction, we get
$t_n=\frac{1}{n!}-\frac{1}{(n+1)!}$
$S_n=\sum _{n=1}^{n=10}{t}_n=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}...\frac{1}{10!}-\frac{1}{11!}$
So all terms will be cancelled except first and last term.

$S_n=\frac{1}{1!}-\frac{1}{11!}$
So, the summation is $\frac{11!-1}{11!}$

Hence, option B.