Sum of Squares of Binomial Coefficients/Inductive Proof

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Theorem

$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$


Proof

For all $n \in \N$, let $\map P n$ be the proposition:

$\ds \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$


$\map P 0$ is true, as this just says:

$\dbinom 0 0^2 = 1 = \dbinom {2 \times 0} 0$

This holds by definition.


Basis for the Induction

$\map P 1$ is true, as this just says:

$\dbinom 1 0^2 + \dbinom 1 1^2 = 1^2 + 1^2 = 2 = \dbinom 2 1$

This also holds by definition.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\ds \sum_{i \mathop = 0}^k \binom k i^2 = \binom {2 k} k$


Then we need to show:

$\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2 = \binom {2 \paren {k + 1} } {k + 1}$


Induction Step

This is our induction step:

\(\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2\) \(=\) \(\ds \binom {k + 1} 0^2 + \sum_{i \mathop = 1}^k \binom {k + 1} i^2 + \binom {k + 1} {k + 1}^2\)
\(\ds \) \(=\) \(\ds 1 + \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} + \binom k i}^2 + 1\) Pascal's Rule
\(\ds \) \(=\) \(\ds 1 + \sum_{i \mathop = 1}^k \paren {\binom k {i - 1}^2 + \binom k i^2 + 2 \binom k {i - 1} \binom k i} + 1\)
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = 0}^{k - 1} \binom k i^2 + 1} + \paren {1 + \sum_{i \mathop = 1}^k \binom k i^2} + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \paren {\sum_{i \mathop = 0}^{k - 1} \binom k i^2 + \binom k k^2} + \paren {\binom k 0^2 + \sum_{i \mathop = 1}^k \binom k i^2} + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^k \binom k i^2 + \sum_{i \mathop = 0}^k \binom k i^2 + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}\)
\(\ds \) \(=\) \(\ds \binom {2 k} k + \binom {2 k} k + 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}\) Induction Hypothesis


Now we look at $\ds 2 \sum_{i \mathop = 1}^k \paren {\binom k {i - 1} \binom k i}$.

Using the Chu-Vandermonde Identity:

$\ds \sum_i \binom r i \binom s {n - i} = \binom {r + s} n$

From the Symmetry Rule for Binomial Coefficients, this can be written:

$\ds \sum_i \binom r i \binom s {s - n + i} = \binom {r + s} n$

Putting $r = k, s = k, s - n = -1$ from whence $n = k + 1$:

$\ds \sum_i \binom k i \binom k {i - 1} = \binom {2 k} {k + 1}$


So:

\(\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i^2\) \(=\) \(\ds 2 \binom {2 k} k + 2 \sum_i \paren {\binom k {i - 1} \binom k i}\) because when $i \le 0$ and $i > k$ we have $\dbinom k {i - 1} \dbinom k i = 0$
\(\ds \) \(=\) \(\ds 2 \binom {2 k} k + 2 \binom {2 k} {k + 1}\)
\(\ds \) \(=\) \(\ds 2 \binom {2 k + 1} {k + 1}\) Pascal's Rule
\(\ds \) \(=\) \(\ds \binom {2 k + 1} k + \binom {2 k + 1} {k + 1}\) Symmetry Rule for Binomial Coefficients
\(\ds \) \(=\) \(\ds \binom {2 k + 2} {k + 1}\) Pascal's Rule
\(\ds \) \(=\) \(\ds \binom {2 \paren {k + 1} } {k + 1}\)


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$

$\blacksquare$

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