Answer
$d^7$
Work Step by Step
Using the $(r+1)$st term of the expansion of $(a+b)^n$, which is given by $\dfrac{n!}{r!(n-r)!}a^{n-r}b^r$, then the $
8
$th term of $
(2c+d)^7
$ is
\begin{array}{l}
\dfrac{7!}{7!0!}(2c)^{0}(d)^{7}
\\\\=
d^7
\end{array}