Question

Find the coefficient of $x^3$ in the expansion of ${(x^2+\frac{3\sqrt{2}}{x})}^9$

Answer 1

using binomial therm

$(x^2+\frac{3\sqrt{2}}{x^9})$ ${\sum }_{n=0}^9$ ${}^9{C}_n$ $\left(x^2{)}^{9-n}\right(\frac{3\sqrt{2}}{x}{)}^n$

${\sum }_{n=0}^9$ ${}^9{C}_n$ $x^{18-2n}$ $\frac{(3\sqrt{2}{)}^n}{x^n}$

$={\sum }_{n=0}^9$ ${}^9{C}_n$ $x^{18-3n}$ $(3\sqrt{2}{)}^n$

$={\sum }_{n=0}^9$ ${}^9{C}_n$ $x^{18-2n}$ $x^{18-2n}$

$T_{n+1}{=}^9{C}_n(3\sqrt{2}{)}^n{x}^{18-3n}$

coefficient of $x^3{=}^9{C}_5(3\sqrt{2}{)}^5{=}^9{C}_5{3}^5.(\sqrt{2}{)}^5[18+3n=33n=18-3n=\frac{15}{3}=5$

$=\frac{9\times 8\times 7\times 6\times 5}{1\times 2\times 3\times 4\times 5}{3}^5.4\sqrt{2}$

$=126.243.4\sqrt{2}.$

$=422,472\sqrt{2}.$