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binomial theorem
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Definition/Summary
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| 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. |
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Equations
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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] |
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Recent forum threads on binomial theorem
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Breakdown
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Mathematics
> Algebra
>> Products
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Extended explanation
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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] |
Commentary
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