Mathematics - Binomial Theorem Question with Solution | TestHub

Given that The coefficient of $x$ and $x^2$ in ${\left(1+x\right)}^p{\left(1-x\right)}^q$ are $4$ and $-5$.

$\Rightarrow {\left(1+x\right)}^p{\left(1-x\right)}^q=\left(1+px+\frac{p\left(p-1\right)}{2!}{x}^2+....\right)\left(1-qx+\frac{q\left(q-1\right)}{2!}{x}^2-....\right)$

Now coefficient of $x$ from the above expansion will be

$p-q$ which is equal to $4$

$\Rightarrow p-q=4$.

Similarly coefficient of $x^2$ is $-5$.

$\Rightarrow \frac{p\left(p-1\right)}{2}+\frac{q\left(q-1\right)}{2}-pq=-5$

$\Rightarrow \frac{p^2-2pq+q^2}{2}-\frac{\left(p+q\right)}{2}=-5$

$\Rightarrow \frac{{\left(p-q\right)}^2}{2}-\frac{\left(p+q\right)}{2}=-5$

$\Rightarrow \frac{16}{2}+5=\frac{\left(p+q\right)}{2}$

$\Rightarrow p+q=26$ and $p-q=4$

On solving the above equations we get,

$p=15$ and $q=11$.

$\Rightarrow 2p+3q=2\left(15\right)+3\left(11\right)=63$

Therefore, the required answer is $63$.