The Binomial Theorem allows us to expand expressions of the form $(a+b)^n$ when $n$ is large. It states, for a positive integer $n$,
\[(a+b)^n = a^n + na^{n-1}b + \frac{n(n-1)}{2!}a^{n-2}b^2 + \frac{n(n-1)(n-2)}{3!}a^{n-3}b^3+ \dotso + b^n\]
A simpler form of the theorem is often quoted, taking the special case $a=1$ and $b=x$. \[(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3+ \dotso + x^n\]
Apply the binomial theorem to expand $(1+x)^3$.
Use the theorem with $n=3$:
\begin{align} (1+x)^3 &=1+3x+\frac{(3)(3-1)}{2!}x^2+\frac{(3)(3-1)(3-2)}{3!}x^3\\ &=1+3x+3x^2+x^3 \end{align}
Find the first three terms in the expansion of $(1+x)^{32}$.
Use the theorem with $n=32$ and stop after 3 terms:
\begin{align} (1+x)^{32} &= 1 +32x +\frac{(32)(32-1)}{2!}x^2 + \dotso\\ &= 1+32x+496x^2+\dotso \end{align}
Find the first four terms in the expansion of $\left(1+\dfrac{1}{3}x\right)^{10}$.
Use the theorem with $n=10$ and replace $x$ with $\dfrac{x}{3}$:
\begin{align} \left(1+\frac{x}{3}\right)^{10} &=1+10\left(\frac{x}{3}\right)+\frac{(10)(10-1)}{2!}\left(\frac{x}{3}\right)^2 + \frac{(10)(10-1)(10-2)}{3!} \left(\frac{x}{3}\right)^3 + \dotso\\ &=1+\frac{10}{2}x+5x^2+\frac{40}{9}x^3+\dotso \end{align}
Find the first three terms in the expansion of $(2-x)^4$.
There are two ways to do this question.
Method 1
Use the $(a+b)^n$ version of the formula, with $a=2$, $b=-x$ and $n=4$:
\begin{align} (2-x)^4 &= 2^4 + (4)(2^3)(-x) + \frac{(4)(4-1)}{2!}(2^2)(-x)^2 + \dotso\\ &=16-32x+24x^2+\dotso \end{align}
Method 2
Use the $(1+x)^n$ version of the formula, by taking a factor of $2^4$ out of the bracket to make the expression into the form $2^4\Bigl(1-\dfrac{x}{2}\Bigr)^4$:
\begin{align} 2^4\left(1-\frac{x}{2}\right)^4 &= 16\Biggl(1+4\left(-\frac{x}{2}\right)+\frac{(4)(4-1)}{2!}\left(-\frac{x}{2}\right)^2\Biggr) +\dotso\\ &=16\left(1-2x+\frac{3x}{2}\right) +\dotso\\ &=16-32x+24x^2+\dotso \end{align}
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.