Question

The coefficient of x and x$^2$ in (1 + x)$^p$ (1 – x)$^q$ are 4 and – 5, then 2p + 3q is

Answer 1

The correct answer is 63
$(1+x)^p(1-x)^q$
$(1+px+\frac{p(p-1)}{2!}x^2+......)$
$(1-qx+\frac{q(q-1)}{2!}x^2-......)$
$p-q=4$
$\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-pq=-5$
p2+q2 - p - 2 - 2pq=-10
(q+4)2 + q2 - (q-4) - q - 2(4+q)q = -10
q2 + 8q + 16 - q2 - q - 4 - q - 8q - 2q2 = -10
-2q=-22
so , q=11 and p=15
$\therefore$ 2p + 3q = 2(15) + 3(11)
= 30 + 33
= 63