Revise The binomial theorem for IB Mathematics: Analysis and Approaches (1.3)

Write out the expansions of $(a+b)^n$ for small $n$ and look only at the numerical coefficients. For $n=0,1,2,3,4$ they are $1$; $\;1\;1$; $\;1\;2\;1$; $\;1\;3\;3\;1$; $\;1\;4\;6\;4\;1$. Stacked up, these rows form Pascal's triangle, and it has a beautifully simple building rule: each entry is the sum of the two entries directly above it, with a $1$ down each edge. The row for exponent $n$ gives every coefficient you need to expand $(a+b)^n$.

$\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \end{array}$

Pascal's triangle is ideal for small powers you can build quickly by hand. For larger $n$ — where writing out every row would be slow — we compute each coefficient directly with the combinations formula instead.

The number in row $n$, position $r$ (counting from $r=0$) is the binomial coefficient $\binom{n}{r}$, which is exactly the ${}^nC_r$ from combinatorics — the number of ways to choose $r$ objects from $n$. That is why it also counts how many times the term $a^{n-r}b^r$ appears when you multiply $(a+b)$ by itself $n$ times. You can evaluate it directly:

$\binom{n}{r}={}^nC_r=\frac{n!}{r!\,(n-r)!}$

Putting the coefficients and the powers together gives the theorem itself. Each term pairs a binomial coefficient with a falling power of $a$ and a rising power of $b$.

$(a+b)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+\binom{n}{n}b^n$

or concisely

$(a+b)^n=\displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r.$

Exam questions often ask for just one term — 'the term in $x^5$', 'the coefficient of $x^2$', or 'the constant term'. Writing out the whole expansion wastes time. Instead, work with the general term, collect the powers of the variable into one exponent, and solve for the $r$ that produces the power you need.

General term: $\;T_{r+1}=\binom{n}{r}a^{n-r}b^r.$

Because the first terms of a binomial expansion are the largest when $b$ is small, the opening terms give a quick numerical approximation. For example, to estimate $(1.02)^6$, write it as $(1+0.02)^6$ and keep only the first few terms: $1+\binom{6}{1}(0.02)+\binom{6}{2}(0.02)^2+\cdots=1+0.12+0.0030+\cdots\approx 1.1230$. Each later term is much smaller than the one before, so a handful of terms already gives several correct decimal places (the exact value is $1.12616\ldots$). Keep more terms for more accuracy.

On Paper 1 the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards the correct approach even if the arithmetic later slips; an accuracy mark (A) is for a correct result and is dependent on the relevant method mark being earned. Follow-through (FT) means a correct answer that flows from an earlier slip can still score, 'ignore subsequent working' (ISW) means correct work is not un-done by later fumbling, and equivalent or exact forms are accepted. Study how each mark below is earned by a specific line.

The binomial theorem reappears throughout the course. The coefficients $\binom{n}{r}$ are the same ${}^nC_r$ that count combinations in probability, and the alternating-sign handling here is the same care you will use expanding brackets in calculus and series work. At HL the theorem extends to any rational power as an infinite series, but the machine you have just built — general term, match the power, substitute — is exactly the routine you carry forward. Master the habit: identify $a,b,n$ with signs, write the general term, fix $r$ from the power you want, then evaluate and show every line.