Question

Find the coefficient of $x^7$ in the expansion of ${(1+3x-2x^3)}^{10}$

Answer 1

Coefficient of $x^7$ in the expansion of ${(1+3x-2x^3)}^{10}$

$=\sum \frac{10!}{\alpha !\beta !\gamma !}{1}^{\alpha }{3}^{\beta }{(-2)}^{\gamma }$

where $\alpha +\beta +\gamma =10$ and $\beta +3\gamma =7$

The possible values of $\alpha ,\beta ,\gamma$ are given below:

$\alpha$	$\beta$	$\gamma$
$3$	$7$	$0$
$5$	$4$	$1$
$7$	$1$	$2$

$\therefore$ Co-efficient of $x^7$

$=\frac{10!}{3!7!0!}{\left(1\right)}^3{\left(3\right)}^7{(-2)}^0+\frac{10!}{5!4!1!}{\left(1\right)}^5{\left(3\right)}^4{(-2)}^1+\frac{10!}{7!1!2!}{\left(1\right)}^7{\left(3\right)}^1{(-2)}^2$ on simplification

$=262440-204120+4320=62640$