If the coefficients of x and x^2 in [(1+x)^p(1-x)^q] are 4 and -5 respectively, then 2 p+3…

Question

If the coefficients of $x$ and $x^{2}$ in $$(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$$ are 4 and -5 respectively, then $2 p+3 q$ is equal to :

Accepted Answer

We have, coefficient of $x$ in $(1+x)^p(1-x)^q=4$ and

coefficient of $x^2$ in $(1+x)^p(1-x)^q=-5$

$$
\begin{aligned}
& (1+x)^p(1-x)^q \\
& =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\
& =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\right) x^2+\ldots \ldots
\end{aligned}
$$

Coefficient of $x$ in $(1+x)^p(1-x)^q=-q+p$

$\Rightarrow p-q=4$ ...........(i)

$$
\text { Coefficient of } x^2 \text { in }(1+x)^p(1-x)^q=\frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2}
$$

$$
\begin{aligned}
& \Rightarrow \frac{q^2-q-2 p q+p^2-p}{2}  =-5 \\
& \Rightarrow \frac{p^2+q^2-2 p q-(p+q)}{2}  =-5 \\
& \Rightarrow (p-q)^2-(p+q)  =-10 \\
& \Rightarrow (4)^2-(p+q)  =-10 \quad[\because \text { From Eq. (i) }] \\
& \Rightarrow p+q  =26 ...........(ii)
\end{aligned}
$$

Form Eqs. (i) and (ii), we get $p=15, q=11$

$\therefore 2 p+3 q=2 \times 15+3 \times 11=30+33=63$

Solutions · 1