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:
Step-by-Step Solution
Key Concept: Coefficient of $x^{k}$ in $(1+x)^{p}(1-x)^{q}$ is $\sum_{i+j=k}\binom{p}{i}\binom{q}{j}(-1)^{j}.$ For $k=1$ and $k=2$, this gives $p-q=1$ and $\binom{p}{2}-pq+\binom{q}{2}=-2.$
Coefficient of $x$: $p-q=1.$
Coefficient of $x^{2}$: $\dfrac{p(p-1)}{2}-pq+\dfrac{q(q-1)}{2}=-2.$ Multiply by $2$:
$p^{2}-p-2pq+q^{2}-q=-4\Rightarrow (p-q)^{2}-(p+q)=-4\Rightarrow 1-(p+q)=-4\Rightarrow p+q=5.$
With $p-q=1$: $p=3,\,q=2.$ Hence $p^{2}+q^{2}=9+4=13.$
Correct Answer: 2