Binomial expansion for positive integer powersEdexcel A-Level Maths: Revision notes

The factorial of a positive integer is $n!=n\times(n-1)\times\ldots\times2\times1$, with $0!=1$. The number of ways to choose $r$ items from $n$ is the binomial coefficient $\binom nr={}^nC_r=\frac{n!}{r!\,(n-r)!}.$ For example $\binom62=\frac{6!}{2!\,4!}=\frac{6\times5}{2}=15$. Your calculator has an $nC_r$ button. Two facts are used all the time: $\binom n0=\binom nn=1$, and symmetry, $\binom nr=\binom{n}{n-r}$. So $\binom64=\binom62=15$.

Pascal's triangle lists the coefficients of $(1+x)^n$. Each entry is the sum of the two above it, so row $5$ is $1,\ 5,\ 10,\ 10,\ 5,\ 1$. This is the rule $\binom nr+\binom n{r+1}=\binom{n+1}{r+1}.$ For example $\binom52+\binom53=10+10=20=\binom63$. The rows are symmetrical because $\binom nr=\binom n{n-r}$. Pascal's triangle is quickest for small $n$ (up to about $6$); use the formula or calculator for larger $n$.

For a positive integer $n$, $(a+bx)^n=a^n+\binom n1a^{n-1}(bx)+\binom n2a^{n-2}(bx)^2+\ldots+(bx)^n.$ The term in $x^r$ is $\binom nra^{n-r}b^rx^r$. The powers of $a$ fall and the powers of $bx$ rise, and the powers in each term add to $n$. Example: $(1+2x)^5=1+5(2x)+10(2x)^2+10(2x)^3+\ldots=1+10x+40x^2+80x^3+\ldots$ When $b$ is negative, the signs alternate: $(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5$. To find an unknown constant, write the required term in terms of it and equate to the given coefficient: for $(2+kx)^5$, the $x^2$ term is $80k^2x^2$.

If $X\sim B(n,p)$ and $q=1-p$, then $P(X=r)=\binom nrp^rq^{n-r}.$ These are exactly the terms of the expansion of $(q+p)^n$, which is why they sum to $1$. For a fair coin tossed $6$ times, $P(X=2)=\binom62\left(\frac12\right)^2\left(\frac12\right)^4=\frac{15}{64}$. The symmetry $\binom nr=\binom n{n-r}$ explains why, when $p=\frac12$, $P(X=r)=P(X=n-r)$.

To expand a product such as $(1+3x)(2-x)^5$, expand the power first, then multiply and collect only the terms you need. For the $x^2$ coefficient, add $1\times80$ and $3\times(-80)$ to get $-160$. To approximate a number, choose $x$ small and write the number in the form of the expression. For $1.02^5$, take $(1+2x)^5$ with $x=0.01$: $1+10(0.01)+40(0.0001)=1.104$. More terms give a closer value. Keep the same $x$ in every factor, and do not forget the numbers multiplying powers of $x$: $1.03\times1.99^5$ comes from $(1+3x)(2-x)^5$ at $x=0.01$.