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

For a positive integer $n$, $n!=n(n-1)(n-2)\cdots2\times1$, with $0!=1$. The binomial coefficient $\binom nr$, also written $^nC_r$, is $\binom nr=\frac{n!}{r!(n-r)!}$ and counts the number of ways of choosing $r$ items from $n$. Useful facts: $\binom n0=\binom nn=1$, $\binom n1=n$, and $\binom nr=\binom n{n-r}$. For example, $\binom62=\frac{6\times5}{2}=15$. The coefficients form Pascal's triangle, where each entry is the sum of the two above it, e.g. row $n=4$ is $1,4,6,4,1$.

For positive integer $n$, $(a+bx)^n=a^n+\binom n1a^{n-1}(bx)+\binom n2a^{n-2}(bx)^2+\cdots+(bx)^n.$ The term in $x^r$ is $\binom nr a^{n-r}b^rx^r$. The powers of $a$ decrease from $n$ to $0$ and the powers of $bx$ increase from $0$ to $n$, and there are $n+1$ terms. Example: $(2+x)^6$ has $x^2$ coefficient $\binom62\times2^4=15\times16=240$ and constant term $2^6=64$.

Put the whole of $bx$ in brackets before raising to a power so that signs and numbers are correct. Example: $(3-2x)^7=3^7+\binom71(3)^6(-2x)+\binom72(3)^5(-2x)^2+\cdots=2187-10206x+20412x^2+\cdots$. Signs alternate when the second term is negative. When $a=1$ the formula simplifies: $(1+3x)^8=1+24x+252x^2+\cdots$ since $\binom82\times9=252$. You can also find $n$ from a given coefficient: if the $x$ term of $(1+3x)^n$ is $24x$ then $3n=24$ and $n=8$.

To find a single coefficient, use the general term $\binom nra^{n-r}b^r$ and do not expand everything. For a product such as $(1+x)(2+x)^6$ or $(2+x)(3-2x)^7$, find the terms in each expansion that multiply to give the power of $x$ you need, and add the products. Example: the $x^2$ coefficient of $(1+x)(2+x)^6$ is $1\times240+1\times192=432$. Approximations: $1.03^8=(1+3\times0.01)^8\approx1+0.24+0.0252=1.2652$ using the first three terms.

If $X\sim B(n,p)$, then $P(X=r)=\binom nrp^r(1-p)^{n-r}$. These are exactly the terms of the expansion of $\left((1-p)+p\right)^n$, which equals $1^n=1$, so the probabilities sum to $1$. Example: $n=5$, $p=\frac14$: $P(X=2)=\binom52\left(\frac14\right)^2\left(\frac34\right)^3=\frac{135}{512}$. Probabilities of the form $P(X\ge r)$ are found by adding the relevant terms; for example $P(X\ge3)=\frac{90+15+1}{1024}=\frac{53}{512}$.