Binomial Theorem Question - None

Question

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

Accepted Answer

$(1+x)^{p}(1-x)^{q}$
$=\left({ }^{p} C_{0}+{ }^{p} C_{1} x+{ }^{p} C_{2} x^{2}+\ldots\right)$
$\left({ }^{9} C_{0}+{ }^{1} C_{1} x+{ }^{9} C_{2} x^{2}+\ldots\right)$
coefficient of $x$ is ${ }^{p} C_{0}{ }^{q} C_{1}+{ }^{p} C_{1}{ }^{q} C_{0}=1$
$p-q=1$
coefficient of $x^{2}$ is
${ }^{p} C_{0}{ }^{q} C_{2}-{ }^{p} C_{1}{ }^{q} C_{1}+{ }^{p} C_{2}{ }^{q} C_{0}=-2$
$\Rightarrow \frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2}=-2$
$\Rightarrow q^{2}-q-2 p q+p^{2}-p=-4$
$\Rightarrow 1-(p+q)=-4[$ since $p-q=1]$
$\Rightarrow p+q=5$
Hence, $p=3, q=2$
$\therefore p^{2}+q^{2}=13$

Question:

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