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 :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Coefficient of $x$: $p-q=1$. Coefficient of $x^2$: $\binom p2+\binom q2-pq=-2$. With $p=q+1$: $\frac{(q+1)q+q(q-1)}2-q(q+1)=q^2-q^2-q=-q=-2$, so $q=2$, $p=3$ and $p^2+q^2=13$. Answer (2).

Q. 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 :

Explanation

Reviewed by Dr. Abubacker

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