What is the Binomial Theorem and How is it Proven?

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Definition/Summary

The binomial theorem gives the expansion of a binomial [itex](x+y)^n[/itex] 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 [itex]n \; \epsilon \; \mathbb{N}[/itex]

[tex](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[/tex]

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


Cases

1. Substituting y=-y we get,

[tex](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[/tex]

2. Having y=1 gives,


[tex](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[/tex]

Extended explanation

Proof by Induction

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

Assuming it to be true for [itex]n=k[/itex]

[tex](x+y)^n = \sum_{r=0}^{k} \binom{n}{k} x^{n-k}y^k[/tex]

Now it needs to hold for [itex]n=k+1[/itex] to complete the inductive step. We use

[tex](x+y)^{k+1} = x(x+y)^k + y(x+y)^k[/tex]

Expanding each [itex](x+y)^k[/itex] individually, multiplying by x and y respectively,

[tex](x+y)^{k+1} = \sum_{r=0}^{k} x^{k+r-1}y^r + \sum_{r=0}^{k} x^{k+r}y^{r+1}[/tex]

Using the property,

[tex]\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}[/tex]

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

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,

[tex](x+y)^n = x^n + nx^{n-1}y+ \frac{n(n-1)}{2}x^{n-2}b^2 + ...+b^n[/tex]

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
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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
 
Greg Bernhardt said:
Generalization

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

(x+y)n=xn+nxn−1y+n(n−1)2xn−2b2+...+bn​
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: [itex]\frac{n\cdot (n-1)\cdot ... \cdot 1}{1\cdot 2 \cdot ... \cdot n}[/itex]. The next one - if n is an integer - will be [itex]\frac{n\cdot (n-1)\cdot ... \cdot 1\cdot 0}{1\cdot 2 \cdot ... \cdot n\cdot (n+1)}=0[/itex].Thus [itex](x+y)^{\alpha}[/itex] with [itex]0<\alpha <1[/itex] will start out as [itex]x^{\alpha}+\alpha x^{\alpha-1}y+...[/itex] where all exponents of x from #2 on will be negative.
 
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