Question

Find the constant term (term independent of x) in the expansion of ${(2x+\frac{1}{{3x}^2})}^9$

Answer 1

Using binomial theorem,

${(2x+\frac{1}{3x^2})}^9=\sum _{n=0}^9{}^9{C}_n(2x{)}^{9.n}{\left(\frac{1}{3x^2}\right)}^n$

$=\sum _{n=0}^9{}^9{C}_n{2}^{9-n}\cdot x^{9-n}\cdot \frac{1}{3^n{x}^{2n}}$

$=\sum _{n=0}^9{}^9{C}_n\frac{2^{9-n}}{3^n}\cdot x^{9-n-2n}$

$=\sum _{n=0}^9{}^9{C}_n\frac{2^{9-n}}{3^n}\cdot x^{9-3n}$

and $T_{r+1}={}^9{C}_r\frac{2^{9-r}}{3^r}{x}^{9-3r}$

Here $9-3r=0$

$\Rightarrow 3r=9$

$\Rightarrow 3r=9$

$\Rightarrow r=3$

$\therefore T_4={}^9{C}_3\frac{2^{9-3}}{3^3}\cdot x^{9-3\left(3\right)}$

$T_4=\frac{9\times 8\times 7}{1\times 2\times 3}\cdot \frac{2^6}{3^3}\cdot x^o$

$=84\cdot \frac{64}{27}$

$=\frac{1792}{9}$

$\therefore$ Coefficient of constant term $=\frac{1792}{9}$.