1 problem found
By considering the coefficient of \(x^{n}\) in the identity \((1-x)^{n}(1+x)^{n}=(1-x^{2})^{n},\) or otherwise, simplify \[ \binom{n}{0}^{2}-\binom{n}{1}^{2}+\binom{n}{2}^{2}-\binom{n}{3}^{2}+\cdots+(-1)^{n}\binom{n}{n}^{2} \] in the cases (i) when \(n\) is even, (ii) when \(n\) is odd.
Solution: The coefficient of \(x^n\) on the LHS is \begin{align*} && (1-x^2)^n &= (1-x)^n(1+x)^n \\ [x^n]: && \begin{cases} (-1)^{\lfloor \frac{n}2 \rfloor}\binom{n}{\lfloor \frac{n}2 \rfloor} &\text{if } n\text{ even} \\ 0 & \text{otherwise} \end{cases} &= \sum_{i=0}^n \underbrace{(-1)^i\binom{n}{i}}_{\text{take }(-x)^i\text{ from first bracket}} \cdot \underbrace{\binom{n}{n-i}}_{\text{take }x^{n-i}\text{ from second bracket}} \\ &&&= \sum_{i=0}^n (-1)^i\binom{n}{i}\binom{n}{i} \\ &&&= \sum_{i=0}^n (-1)^i\binom{n}{i}^2\\ \end{align*}