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

$n!=n(n-1)(n-2)\cdots2\cdot1$ with $0!=1$. The binomial coefficient is $\binom nr={}^nC_r=\frac{n!}{r!\,(n-r)!},$ the number of ways to choose $r$ items from $n$. It is symmetrical, $\binom nr=\binom n{n-r}$, with $\binom n0=\binom nn=1$ and $\binom n1=n$. Example: $\binom83=\frac{8\times7\times6}{3\times2\times1}=56$. The coefficients also form Pascal's triangle: each entry is the sum of the two above it.

For a positive integer $n$: $(a+bx)^n=\sum_{r=0}^{n}\binom nr a^{n-r}(bx)^r=a^n+\binom n1a^{n-1}bx+\binom n2a^{n-2}b^2x^2+\dots+b^nx^n.$ The powers of $a$ fall from $n$ to $0$ while the powers of $bx$ rise from $0$ to $n$; the powers in each term add up to $n$. There are $n+1$ terms. Example: $(2+x)^6=64+192x+240x^2+160x^3+60x^4+12x^5+x^6$.

Put each part of the bracket in brackets before raising to a power, especially negatives and fractions. Expand $(3-2x)^4$: $81+4(27)(-2x)+6(9)(4x^2)+4(3)(-8x^3)+16x^4=81-216x+216x^2-96x^3+16x^4.$ The signs alternate when $b$ is negative. If the bracket is $(1+kx)^n$ the coefficients are $\binom nrk^r$; for example $\binom82k^2=252$ gives $28k^2=252$ and $k=3$.

The term in $x^r$ is $\binom nr a^{n-r}b^rx^r$, so you can write down one coefficient without the full expansion. Example: the coefficient of $x^3$ in $(2+x)^6$ is $\binom63\times2^3=160$. If a bracket multiplies the expansion, collect the products that give the required power. For $(2-x)(1+3x)^8$ the coefficient of $x^3$ is $2\times\binom83 3^3-1\times\binom823^2=3024-252=2772$. Unknown constants come from equating a coefficient to a given value, then solving.

To estimate a number such as $2.98^4$, write it as $(3-2x)^4$ with $x=0.01$ and substitute into the first few terms: $81-2.16+0.0216-0.000096\approx78.8615$. The terms get small quickly because $x$ is small, so a few terms give a good estimate.