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A binomial squared sum

Prove that

\displaystyle \binom{n}{0}^2 + \binom{n}{1} ^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}

Solution

Let us begin by recalling the binomial expansion

\displaystyle \left ( 1+x \right )^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} x^k

However,

    \begin{align*} \left ( 1+x \right )^{2n} &= \left ( 1+x \right )^n \left ( 1+x \right )^n \\ &= \left [ \sum_{i=0}^{n} \binom{n}{i} x^i \right ] \left [ \sum_{j=0}^{n} \binom{n}{j} x^j \right ]\\ &=\sum_{i, j} \binom{n}{i} \binom{n}{j} x^{i+j} \\ &= \sum_{k=0}^{2n} x^k \sum_{i+j=k}\binom{n}{i} \binom{n}{j} \end{align*}

Equatating the coefficients of x^n we have that

    \begin{align*} \binom{2n}{n} &= \sum_{i+j=n} \binom{n}{i} \binom{n}{j} \\ &= \sum_{i=0}^{n} \binom{n}{i} \binom{n}{n-i}\\ &= \sum_{i=0}^{n} \binom{n}{i}^2 \end{align*}

yielding the result.

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