Answer
$7x^5$
Work Step by Step
Our aim is to find the fourth term for $(x+\dfrac{1}{2})^{8}$.
General formula:$(p+q)^n=\displaystyle \binom{n}{k}p^{n-k}q^k$
and $\displaystyle \binom{n}{k}=\dfrac{n!}{k!(n-k)!}$
Now, $(x+\dfrac{1}{2})^{8}=\displaystyle \binom{8}{3}(x)^{8-3}(\dfrac{1}{2})^3$
This implies,
$=\dfrac{8!}{3!(8-3)!}(x)^{5}(\dfrac{1}{2})^3$
$=\dfrac{ 8 \times 7 \times 6 \times 5!}{(3 \times 2 \times 1)5!}x^{5}(\dfrac{1}{8})$
Thus,
$=7x^5$
