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

Question

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

Accepted Answer

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

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

$\begin{gathered} (1+x{)}^{p}(1-x{)}^{q} \ =(1+px+\frac{p(p-1)}{2}{x}^{2}+\ldots )(1-qx+\frac{q(q-1)}{2}{x}^{2}+\ldots ) \ =1+(p-q)x+(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq){x}^{2}+\ldots \ldots \end{gathered}$

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

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

${x}^{2}(1+x{)}^{p}(1-x{)}^{q}=\frac{q(q-1)}{2}-pq+\frac{p(p-1)}{2}$

$\begin{gathered} \Rightarrow \frac{{q}^{2}-q-2pq+{p}^{2}-p}{2}=-5 \ \Rightarrow \frac{{p}^{2}+{q}^{2}-2pq-(p+q)}{2}=-5 \ \Rightarrow (p-q{)}^{2}-(p+q)=-10 \ \Rightarrow (4{)}^{2}-(p+q)=-10\, [\because ] \ \Rightarrow p+q=26...........(ii) \end{gathered}$

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

$	herefore 2p+3q=2	imes 15+3	imes 11=30+33=63$

Solutions · 1