Question
Let $n$ be a positive integer. Prove that$$\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{l}n \\k\end{array}\right)^{2}=\left\{\begin{array}{ll}0 & \text { if } n \text { is odd } \\(-1)^{m}\left(\begin{array}{c}2 m \\m\end{array}\right) & \text { if } n=2 m .\end{array}\right.$$(Hint: For $n=2 m$, consider the coefficient of $x^{n}$ in $\left(1-x^{2}\right)^{n}=(1+x)^{n}(1-x)^{n}$.)
Step 1
We can write the sum as $$ \sum_{k=0}^{n}(-1)^{k}\binom{n}{k}^2 = \binom{n}{0}^2 - \binom{n}{1}^2 + \binom{n}{2}^2 - \cdots + (-1)^n \binom{n}{n}^2. $$ Now, notice that $\binom{n}{k} = \binom{n}{n-k}$, so we can rewrite the sum Show more…
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(a) Prove that $$\sum_{k=0}^{l}\left(\begin{array}{l}n \\k\end{array}\right)\left(\begin{array}{c}m \\l-k \end{array}\right)=\left(\begin{array}{c}n+m \\l\end{array}\right)$$ Hint: Apply the binomial theorem to $(1+x)^{n}(1+x)^{n}$ (b) Prove that $$ \sum_{i=0}^{n}\left(\begin{array}{l}n \\k\end{array}\right)^{2}=\left(\begin{array}{l}2 n \\n \end{array}\right)$$.
Prove that the following are equivalent for the integer $m$ : (a) $n$ is odd. (b) There exists $k \in \mathbf{Z}$ such that $n=2 k-1$. (c) $n^{2}+1$ is even.
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