## Intermediate Algebra for College Students (7th Edition)

$110$
Since, we have $\sum_{i=1}^{5}[\dfrac{(i+2)!}{i!}]=\dfrac{3!}{1!}+\dfrac{4!}{2!}+\dfrac{5!}{3!}+\dfrac{6!}{4!}+\dfrac{7!}{5!}$ $=6+12+20+30+42$ Hence, we get or, $=110$