Question

If in a series $t_n=\frac{n}{(n+1)!}$ then ${\sum }_{n=1}^{20}{t}_n$ is equal to

• A. $\frac{20!-1}{20!}$
• B. $\frac{21!-1}{21!}$
• C. $\frac{1}{2(n-1)!}$
• D. none of these

Answer 1

The correct option is B $\frac{21!-1}{21!}$

$t_n=\frac{(n+1)-1}{(n+1)!}$
$=\frac{1}{n!}-\frac{1}{(n+1)!}$
$\therefore S_n={\sum }_{n=1}^{20}{t}_n=(\frac{1}{1!}-\frac{1}{2!})+(\frac{1}{2!}-\frac{1}{3!})+(\frac{1}{3!}-\frac{1}{4!})+...+(\frac{1}{19!}-\frac{1}{20!})+(\frac{1}{20!}-\frac{1}{21!})$
$=1-\frac{1}{21!}$
Hence, Option B is correct