Question

Find the coefficient of $x^5$ in $(x+3{)}^8$

Answer 1

It is known that $(r+1)$th term $\left(T_{r+1}\right)$ in the binomial expansion of
$(a+b{)}^n$ is given by $T_{r+1}{=}^n{C}_r{a}^{a-r}{b}^r$
Assuming that $x^5$ occurs in the $(r+1)$th term of expansion $(x+3{)}^8$ we obtain
$T_{r+1}{=}^8{C}_r\left(x{)}^{8-r}\right(3{)}^r$
Comparing the indices of $x$ in $x^5$ and in $T_{r+1}$, we obtain $r=3$
Thus, the coefficient of $x^5$ is
${}^8{C}_3(3{)}^3=\frac{8!}{3!5!}x{3}^3=\frac{\mathrm{8.7.6.5}!}{\mathrm{3.2.5}!}{.3}^3=1512$