Binomial Expansion
Expanding (a + b)ⁿ without slog — Pascal's triangle hands you the coefficients, and the nCr formula lets you reach straight for any single term you need.
Expanding $(a+b)^2$ by hand is easy; $(a+b)^3$ is tedious; $(a+b)^{10}$ by repeated multiplication is a nightmare. The binomial expansion is the shortcut — a formula that writes down every term of $(a+b)^n$ at once. But rather than memorise it cold, it's worth seeing where the numbers come from, because then the formula becomes obvious rather than magic.
Multiply $(a+b)$ by itself a few times and watch the coefficients:
$(a+b)^1 = 1a + 1b$
$(a+b)^2 = 1a^2 + 2ab + 1b^2$
$(a+b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3$
The coefficients — $1,1$ then $1,2,1$ then $1,3,3,1$ — are exactly the rows of Pascal's triangle. That's not a coincidence; it's the whole engine of the expansion.
When you multiply $(a+b)^{n-1}$ by one more $(a+b)$, every term gets hit by both an $a$ and a $b$ — so each coefficient in the new row is fed by two coefficients from the row above, added together. That "sum of the two above" rule is the multiplication, written as a picture. So the coefficients of $(a+b)^n$ are row $n$ of Pascal's triangle, and each one counts how many ways you can pick $b$ from the $n$ brackets — which is exactly what $\binom{n}{r}$ means.
The binomial formula
Reading the coefficients as "choose" numbers gives the general expansion:
$(a+b)^n = \displaystyle\binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n$
where $\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ is the number of ways to choose $r$ items from $n$ (the button labelled $^nC_r$ on your calculator). Two patterns make every term easy to write: the power of $a$ counts down from $n$ to $0$, while the power of $b$ counts up from $0$ to $n$, and the two powers in each term always add to $n$.
To find just one term (say "the $x^5$ term") without expanding everything, use the general term $\dbinom{n}{r}a^{n-r}b^r$. Work out which $r$ gives the power you want, then evaluate only that term. This is far faster than writing the whole expansion, and it's what most exam questions actually ask for.
Worked example 1
Don't forget the powers of the 2 — a very common slip is writing $1+4x+6x^2+\ldots$ as if $a=1$. Every term carries both a Pascal coefficient and a power of the number $2$. Check the ends: the first term is $2^4=16$, the last is $x^4$ — if those are right, the structure is usually sound.
Worked example 2
Pascal's triangle to row 10 is slow to write, so use the $\binom{n}{r}$ formula for just the terms we need.
The step everyone rushes is raising the whole $b$ term to the power: $(3x)^2$ is $9x^2$, not $3x^2$ — the coefficient $3$ gets squared too. Bracket the entire term $(3x)^r$ before evaluating and this never goes wrong. With $a=1$ the expansion collapses to just $\binom{n}{r}(3x)^r$, which is why the "$1+\ldots$" form is so common in exams.
Worked example 3
We only want one term, so go straight to the general term $\binom{n}{r}a^{n-r}b^r$ and find the $r$ that produces $x^3$.
The general term revealed something a blind expansion would have hidden: the power of $x$ is always $6-2r$, which is even for every $r$. So this expansion contains only even powers of $x$, and asking for $x^3$ has the clean answer "coefficient $0$". Setting the power equal to what you want — and checking the solution is a valid whole number $0 \le r \le n$ — is the real skill in "find the term" questions.
When the term being raised has a coefficient — like $(3x)^r$ or $(2x)^{n-r}$ — the whole bracket is raised to the power, coefficient included: $(2x)^3 = 8x^3$, not $2x^3$. Always write the term in brackets first, then apply the power to everything inside. Dropping the coefficient's power is the single most frequent error in this topic.
Check yourself
Find the first three terms, in ascending powers of $x$, of the expansion of $(1 - 2x)^{8}$.
(Hint: with $a=1$, each term is $\binom{8}{r}(-2x)^r$ — remember to raise the whole $(-2x)$ to the power.)
The key detail is $(-2x)^2 = +4x^2$ — squaring the negative makes it positive, so the third term is $+112x^2$. Raising the whole $(-2x)$ each time also produces the alternating signs.