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Binomial expansion

It is well-known that $$(x+y)^2=x^2+2xy+y^2$$. The binomial expansion is a generalization of this formula for other exponents.

The binomial expansion states that, for any integer $$n\ge0$$, $$$ (a+b)^n = a^n+\binom{n}{1} a^{n-1}b + \dots +\binom{n}{n-1} a b^{n-1} + b^n. $$$

The binomial coefficient $$\binom{n}{p}$$ (pronounced "$$n$$ choose $$p$$") is: $$$ \binom{n}{p} = \frac{n!}{p!(n-p)!} = \frac{n\times(n-1)\times\dots\times(n-p+1)} { p\times(p-1)\times\dots\times 1} $$$

The factorial is $$n!=1\times2\times \dots\times n$$, with convention $$0!=1$$.

For $$n=4$$, we get $$$ (a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4$$$

The binomial expansion remains valid for any number $$n$$ (real number, positive or negative) provided $$\vert b/a\vert\lt 1$$ $$$ (a+b)^n = a^n + n a^{n-1}b + \dots +\frac{n(n-1)\dots(n-k+1)}{k!} a^{n-k} b^{k} +\dots.$$$ It is then an infinite expansion, which is convergent.

It is often written for $$\vert x\vert\lt 1$$ as $$$ (1+x)^n = 1 + n x + \dots +\frac{n(n-1)\dots(n-k+1)}{k!} x^k +\dots.$$$

The infinite expression is an asymptotic expansion. The expansion is given up a certain power of $$x$$, the order of the expansion.

The expansion of order $$2$$ of $$(1+x)^{-2}$$ is $$$(1+x)^{-2}=1-2x + \frac{(-2)(-2-1)}{2!}x^2+\dots = 1-2x+3x^2+\dots$$$

The binomial coefficient satisfies the formula $$$ \binom{n+1}{p+1} = \binom{n}{p} + \binom{n}{p+1} $$$

Pascal's triangle is a simple tool for finding binomial coefficients - the entry on the $$n^{\text{th}}$$ row in the $$r^{\text{th}}$$ place is $$\displaystyle\binom{n}{r}$$. The numbers on the left and the right are always $$1$$. The rest of the entries are the sum of the two numbers directly above.

From Pascal's triangle, we see that $$\displaystyle\binom{5}{3} = 10$$.

Pascal's triangle allows us to find binomial coefficients
Pascal's triangle allows us to find binomial coefficients

The binomial identity is proved by direct computation. \begin{align*} &\binom{n}{p} + \binom{n}{p+1} = \frac{n!}{p! (n-p)!} + \frac{n!}{(p+1)! (n-p-1)!}\\ &\quad = \frac{n!}{p! (n-p-1)!}\big(\frac{1}{n-p}+\frac{1}{p+1}\big) \\ &\quad = \frac{n!}{p! (n-p-1)!}\frac{n+1}{\big((n+1)-(p+1)\big)(p+1)}\\ &\quad = \binom{n+1}{p+1}. \end{align*}

The binomial expansion can be proved by induction from the binomial identity $$$ \binom{n+1}{p+1} = \binom{n}{p} + \binom{n}{p+1} $$$

The binomial expansion is obvious when $$n=0$$ (as it reads $$1=1$$).

Assuming it is correct at $$n$$, we find \begin{align*} (a+b)^{n+1} & = (a+b)^n(a+b)\\ & = \left(\sum_{p=0}^n \binom{n}{p} a^{n-p}b^p \right)(a+b)\\ & = \sum_{p=0}^n \binom{n}{p} a^{n-p+1} b^p + \sum_{p=0}^n \binom{n}{p} a^{n-p} b^{p+1}\\ & = a^{n+1} + b^{n+1} + \sum_{p=0}^{n-1} \big(\binom{n}{p+1}+ \binom{n}{p}\big) a^{n-p}b^{p+1}\\ & = a^{n+1} + b^{n+1} + \sum_{p=0}^{n-1} \binom{n+1}{p+1} a^{n-p}b^{p+1}\\ & =\sum_{p=0}^{n+1} \binom{n+1}{p} a^{n+1-p}b^{p} \end{align*} This gives the binomial expansion at rank $$n+1$$.