Sum of Squares of Binomial Coefficients

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Theorem

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

where $\dbinom n i$ denotes a binomial coefficient.


Combinatorial Proof

Consider the number of paths in the integer lattice from $\tuple {0, 0}$ to $\tuple {n, n}$ using only single steps of the form:

$\tuple {i, j} \to \tuple {i + 1, j}$
$\tuple {i, j} \to \tuple {i, j + 1}$

that is, either to the right or up.

This process takes $2 n$ steps, of which $n$ are steps to the right.

Thus the total number of paths through the graph is equal to $\dbinom {2 n} n$.


Now let us count the paths through the grid by first counting the paths:

$(1): \quad$ from $\tuple {0, 0}$ to $\tuple {k, n - k}$

and then the paths:

$(2): \quad$ from $\tuple {k, n - k}$ to $\tuple {n, n}$.

Note that each of these paths is of length $n$.


Since each path is $n$ steps long, every endpoint will be of the form $\tuple {k, n - k}$ for some $k \in \set {1, 2, \ldots, n}$, representing $k$ steps right and $n-k$ steps up.

Note that the number of paths through $\tuple {k, n - k}$ is equal to $\dbinom n k$, since we are free to choose the $k$ steps right in any order.


We can also count the number of $n$-step paths from the point $\tuple {k, n - k}$ to $\tuple {n, n}$.

These paths will be composed of $n - k$ steps to the right and $k$ steps up.

Therefore the number of these paths is equal to $\dbinom n {n - k} = \dbinom n k$.


Thus the total number of paths from $\tuple {0, 0}$ to $\tuple {n, n}$ that pass through $\tuple {k, n - k}$ is equal to the product of:

the number of possible paths from $\tuple {0, 0}$ to $\tuple {k, n - k}$, which equals $\dbinom n k$

and:

the number of possible paths from $\tuple {k, n - k}$ to $\tuple {n, n}$, which equals $\dbinom n k$.

So the total number of paths through $\tuple {k, n - k}$ is equal to $\dbinom n k^2$.


Summing over all possible values of $k \in 0, \ldots, n$ gives the total number of paths.

Thus we get:

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

$\blacksquare$


Inductive 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:

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

$\blacksquare$


Algebraic Proof

Consider the Binomial Theorem:

\(\ds \forall n \in \Z_{\ge 0}: \, \) \(\ds \paren {1 + x}^n\) \(=\) \(\ds \sum_{j \mathop = 0}^n \dbinom n j x^j\)
\(\ds \) \(=\) \(\ds \dbinom n 0 x^0 + \dbinom n 1 x^1 + \dbinom n 2 x^2 + \cdots + \dbinom n {n - 1} x^{n - 1} + \dbinom n n x^n\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \dbinom n n x^0 + \dbinom n {n - 1} x^1 + \dbinom n {n - 2} x^2 + \cdots + \dbinom n 1 x^{n - 1} + \dbinom n 0 x^n\) Symmetry Rule for Binomial Coefficients


Let $m$ be the coefficient of $x^n$ in the expansion of $\paren {1 + x}^{2 n} = \paren {1 + x}^n \times \paren {1 + x}^n$.

For all $k \in \set {0, 1, \ldots, n}$, by matching up the coefficient of $x^k$ with that of $x^{n - k}$ in $(1)$, we have:

$m = \ds \sum_{k \mathop = 0}^n \dbinom n k^2$

But from the Binomial Theorem, this given by:

$m = \dbinom {2 n} n$

Hence the result.


Sources