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

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Aicharya · Accepted Answer

$\begin{aligned} & (1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{\mathrm{q}}=\left({ }^{\mathrm{p}} \mathrm{C}_0+{ }^{\mathrm{p}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_2 \mathrm{x}^2+\ldotsight)\left({ }^{\mathrm{q}} \mathrm{C}_0-{ }^{\mathrm{q}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{q}} \mathrm{C}_2 \mathrm{x}^2+\ldotsight) \ & 	ext {coeff of } \mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_1{ }^{\mathrm{q}} \mathrm{C}_0=1 \ & \mathrm{p}-\mathrm{q}=1 \ & 	ext {coeff of } \mathrm{x}^2 \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_2-{ }^{\mathrm{p}} \mathrm{C}_1{ }^{\mathrm{q}} \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_2{ }^{\mathrm{q}} \mathrm{C}_0=-2 \ & \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2 \ & \mathrm{q}^2-\mathrm{q}-2 \mathrm{pq}+\mathrm{p}^2-\mathrm{p}=-4 \ & 1-(\mathrm{p}+\mathrm{q})=-4 \ & \mathrm{p}+\mathrm{q}=5 \ & \mathrm{p}=3 \ & \mathrm{q}=2 \ & 	ext {so } \mathrm{p}^2+\mathrm{q}^2=13\end{aligned}$

If in the expansion of $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$, the coefficients of $x$ and $x^2$ are 1 and…

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