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

Given that The coefficient of $x$x and ${x}^{2}$x2 in ${(1+x)}^{p}{(1-x)}^{q}$(1+x)p(1−x)q are $4$4 and $-5$−5.

$$\Rightarrow {(1+x)}^{p}{(1-x)}^{q}=(1+px+\frac{p(p-1)}{2!}{x}^{2}+....)(1-qx+\frac{q(q-1)}{2!}{x}^{2}-....)$$⇒(1+x)p(1−x)q=(1+px+2!p(p−1)​x2+....)(1−qx+2!q(q−1)​x2−....)

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

$p-q$p−q which is equal to $4$4

$\Rightarrow p-q=4$⇒p−q=4.

Similarly coefficient of ${x}^{2}$x2 is $-5$−5.

$$\Rightarrow \frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5$$⇒2p(p−1)​+2q(q−1)​−pq=−5

$$\Rightarrow \frac{{p}^{2}-2pq+{q}^{2}}{2}-\frac{(p+q)}{2}=-5$$⇒2p2−2pq+q2​−2(p+q)​=−5

$$\Rightarrow \frac{{(p-q)}^{2}}{2}-\frac{(p+q)}{2}=-5$$⇒2(p−q)2​−2(p+q)​=−5

$$\Rightarrow \frac{16}{2}+5=\frac{(p+q)}{2}$$⇒216​+5=2(p+q)​

$\Rightarrow p+q=26$⇒p+q=26 and $p-q=4$p−q=4

On solving the above equations we get,

$p=15$p=15 and $q=11$q=11.

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

Therefore, the required answer is $63$63.