Menu Home Action My entries Defined browse Select Select in the list MathematicsPhysics Then Select Select in the list Then Select Select in the list Search

binomial theorem

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

 Scientists Blaise Pascal (1623-1662)

 Recent forum threads on binomial theorem

 Breakdown Mathematics > Algebra >> Products

 Images

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

Commentary

 Infinitum @ 05:14 AM May30-12 Recreated as per tiny-tim's suggestion. Added some more stuff, too.