Question

The coefficient of $x^5$ in the expansion of ${(1+3x-2x^3)}^{10}$ is............

Answer 1

$f\left(x\right)={(1+3x-2x^3)}^{10}$

Applying Multinomial expansion

General Term is $=\frac{10!}{a!b!c!}{1}^a{\left(3x\right)}^b{(-2x^3)}^c$ Where $a+b+c=10$

So for $x^5$

We got $b+3c=5$

So possible values of are

$\left(1\right)c=0,b=5$ so$a=5$

$\left(2\right)c=1,b=2$ so$a=7$

So the coefficient of $x^5=\frac{10!}{5!5!0!}{3}^5{(-2)}^0+\frac{10!}{7!2!1!}{3}^2{(-2)}^1$

Coefficient $=61236-6480=54756$