What is the Binomial Theorem and How is it Proven?

In summary, the binomial theorem gives the expansion of a binomial (x+y)^n as a summation of terms. It is closely related to Pascal's triangle and can be written in both equation and summation form. It can also be generalized to any value of 'n', even if it is not a positive integer, but in those cases the expansion will generally not stop. The proof of the theorem uses induction and the generalization formula shows that the terms will have negative exponents if n is not an integer.
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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
 
  • #3
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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