Exercise 1.4.24* (Conundrum 6/8)

Question

Show that
$$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^2 = (-1)^n \binom{2n}{n}.$$

Richard Ganaye · Accepted Answer

\begin{proof} We compare the coefficient $a_{2n}$ of $x^{2n}$ in two expansions of the polynomial $$f(x) = (1-x^2)^{2n} = \sum_{k=0}^{4n} a_ix^i.$$

First
\begin{align*}
(1-x^2)^{2n} &= \sum_{l=0}^{2n} (-1)^l \binom{2n}{l} x^{2l},
\end{align*}
thus $$a_{2n} = (-1)^n \binom{2n}{n}.$$
Next
\begin{align*}
(1-x^2)^{2n} &= (1-x)^{2n}(1+x)^{2n}\
&=\left[\sum_{k=0}^{2n} (-1)^k \binom{2n}{k} x^kight] \left[ \sum_{j=0}^{2n} \binom{2n}{j}x^j ight]\
&= \sum_{l=0}^{4n} \left [\sum_{k=0}^l (-1)^k \binom{2n}{k}\binom{2n}{l-k}ight] x^l\
\end{align*}\
Thus 
\begin{align*}
a_{2n} &= \sum_{k=0}^{2n} (-1)^k \binom{2n}{k}\binom{2n}{2n-k}\
&= \sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^2
\end{align*}
The comparison of the two expressions of $a_{2n}$ gives
$$\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^2 = (-1)^n \binom{2n}{n}.$$

\end{proof}

Exercise 1.4.24* (Conundrum 6/8)

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Exercise 1.4.24* (Conundrum 6/8)

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