Answer
$7x^5$
Work Step by Step
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,$=7x^5$
