Answer
$110$
Work Step by Step
Here, we have $\sum_{i=1}^5 \dfrac{(i+2)!}{i!}= \dfrac{(1+2)!}{1!}+ \dfrac{(2+2)!}{2!}+ \dfrac{(3+2)!}{3!}+ \dfrac{(4+2)!}{4!}+ \dfrac{(5+2)!}{5!}$
$=\dfrac{3!}{1!}+\dfrac{4!}{2!}+\dfrac{5!}{3!}+\dfrac{6!}{4!}+\dfrac{7!}{5!}$
$=\dfrac{6}{1}+\dfrac{24}{2}+\dfrac{120}{6}+\dfrac{720}{24}+\dfrac{5040}{120}$
$=110$