# What is binomial theorem

1. Jul 23, 2014

### Greg Bernhardt

Definition/Summary

The binomial theorem gives the expansion of a binomial $(x+y)^n$ as a summation of terms. The binomial theorem for positive integral values of 'n', is closely related to Pascal's triangle.

Equations

The theorem states, for any $n \; \epsilon \; \mathbb{N}$

$$(x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 +......+\binom{n}{n}x^0y^n$$

In summation form,
$$(x+y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r}y^r$$

Cases

1. Substituting y=-y we get,

$$(x-y)^n = \binom{n}{0}x^ny^0 - \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 -......+ (-1)^{n}\binom{n}{n}x^0y^n$$

2. Having y=1 gives,

$$(x+1)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2} +......+ \binom{n}{n}x^0$$

Extended explanation

Proof by Induction

When $n=0$, the statement obviously holds true, giving $(x+y)^0= \binom{0}{0}=1$

Assuming it to be true for $n=k$

$$(x+y)^n = \sum_{r=0}^{k} \binom{n}{k} x^{n-k}y^k$$

Now it needs to hold for $n=k+1$ to complete the inductive step. We use

$$(x+y)^{k+1} = x(x+y)^k + y(x+y)^k$$

Expanding each $(x+y)^k$ individually, multiplying by x and y respectively,

$$(x+y)^{k+1} = \sum_{r=0}^{k} x^{k+r-1}y^r + \sum_{r=0}^{k} x^{k+r}y^{r+1}$$

Using the property,

$$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$$

We get,
$$(x+y)^{k+1} = \sum_{r=0}^{k+1} \binom{k}{r} x^{(k+1)-r}y^r$$

This completes our inductive step, proving the theorem.

Generalization

For any value of 'n', whether positive, negative, or fractional, the binomial expansion is given by,

$$(x+y)^n = x^n + nx^{n-1}y+ \frac{n(n-1)}{2}x^{n-2}b^2 + ......+b^n$$

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2. Sep 29, 2017

### JBD

Sir, in the generalization part, I have found some sources where the terms are infinite and not finite as written here. Can you post a link where I can find this formula? Thanks

3. Sep 29, 2017

### Svein

If n is not an integer, the binomial expansion will generally not stop. The reason why it stops when n is an integer - look at the (n+1)'th coefficient: $\frac{n\cdot (n-1)\cdot ... \cdot 1}{1\cdot 2 \cdot ... \cdot n}$. The next one - if n is an integer - will be $\frac{n\cdot (n-1)\cdot ... \cdot 1\cdot 0}{1\cdot 2 \cdot ... \cdot n\cdot (n+1)}=0$.Thus $(x+y)^{\alpha}$ with $0<\alpha <1$ will start out as $x^{\alpha}+\alpha x^{\alpha-1}y+...$ where all exponents of x from #2 on will be negative.