## Precalculus (6th Edition) Blitzer

$7x^5$
Calculate the fourth term for $(x+\dfrac{1}{2})^{8}$ by using General formula such as:$(m+n)^r=\displaystyle \binom{r}{k}m^{r-k}n^k$ and $\displaystyle \binom{r}{k}=\dfrac{r!}{k!(r-k)!}$ This implies,$(x+\dfrac{1}{2})^{8}=\displaystyle \binom{8}{3}x^{(8-3)}(\dfrac{1}{2})^3$ $=(\dfrac{1}{8})[\dfrac{8!}{3!(8-3)!}]x^{5}$ $=(\dfrac{1}{8})[\dfrac{ 8 \times 7 \times 6 \times 5!}{(3 \times 2 \times 1)5!}]x^{5}$ or,$=7x^5$