Sum of Squares of Binomial Coefficients
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: $3.12$