Question

The value of $\left(\sum _{k=0}^{100}\frac{k+3}{(k+1)!+(k+2)!+(k+3)!}\right)+\frac{1}{(103)!}$ is equal to

• A. 1
• B. $\frac{1}{2}$
• C. 100
• D. 101

Answer 1

Answer: (B)

$\sum _{k=0}^{100}\left(\frac{k+3}{(k+1)!+(k+2)!+(k+3)!}\right)=\sum _{k=0}^{100}\frac{(k+3)}{(k+1)![1+k+2+(k+3)(k+2)]}$
$=\sum _{k=0}^{100}\frac{k+3}{(k+1)!\left(k^2+6k+9\right)}=\sum \frac{k+3}{(k+1)(k+3{)}^2}=\sum _{k=0}^{100}\frac{1}{(k+3)(k+1)!}$
$=\sum _{k=0}^{100}\frac{k+2=(k+3)-1}{(k+3)!}=\sum _{k=0}^{100}\frac{1}{(k+2)!}-\frac{1}{(k+3)!}=\frac{1}{2!}-\frac{1}{(103)!}$
$\therefore S=\frac{1}{2}\Rightarrow \text{ B}$