Question

If $a_n=\frac{n}{\left(n+1\right)!},$ then $\sum _{n=1}^{50}{a}_n$ is equal to

• A. $\frac{50!-1}{50!}$
• B. $\frac{51!-1}{51!}$
• C. $\frac{1}{2\left(n-1\right)!}$
• D. none of these

Answer 1

The correct option is B $\frac{51!-1}{51!}$

$\sum _{n=1}^{50}{t}_n$

$=\sum _{n=1}^{50}\frac{n}{(n+1)!}$

$=\sum _{n=1}^{50}\frac{n+1-1}{(n+1)!}$

$=\sum _{n=1}^{50}\frac{n+1}{(n+1)!}-\sum _{n=1}^{50}\frac{1}{(n+1)!}$

$=\sum _{n=1}^{50}\frac{n+1}{(n+1)n!}-\sum _{n=1}^{50}\frac{1}{(n+1)!}$

$=\sum _{n=1}^{50}\frac{1}{n!}-\sum _{n=1}^{50}\frac{1}{(n+1)!}$

$=\frac{1}{1}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{49!}-\frac{1}{50!}+\frac{1}{50!}-\frac{1}{51!}$

$=1-\frac{1}{51!}$

$=\frac{51!-1}{51!}$