If the coefficients of x and x^2 in the expan | vedclass

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(A) The expansion is given by $(1+x)^p(1-x)^q = (1+px+\frac{p(p-1)}{2}x^2+\dots)(1-qx+\frac{q(q-1)}{2}x^2-\dots)$.The coefficient of $x$ is $p-q=4$,so

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(A) The expansion is given by $(1+x)^p(1-x)^q = (1+px+\frac{p(p-1)}{2}x^2+\dots)(1-qx+\frac{q(q-1)}{2}x^2-\dots)$.The coefficient of $x$ is $p-q=4$,so $p=q+4$.The coefficient of $x^2$ is $\frac{p(p-1)}{2} + \frac{q(q-1)}{2} - pq = -5$.Multiplying by $2$,we get $p^2-p+q^2-q-2pq = -10$,which simplifies to $(p-q)^2 - (p+q) = -10$.Substituting $p-q=4$,we get $4^2 - (p+q) = -10$,so $16 - (p+q) = -10$,which gives $p+q = 26$.Solving $p-q=4$ and $p+q=26$,we add the equations to get $2p=30$,so $p=15$.Then $q = 26-15 = 11$.Finally,$2p+3q = 2(15)+3(11) = 30+33 = 63$.

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

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