What the theorem is doing

A binomial is an algebraic expression made by adding or subtracting two terms. The binomial theorem gives a general way to expand powers of a binomial, so you don’t have to multiply out one bracket after another.

For a positive integer power, the expansion is

Written out, this says

$$(a+b)^n=a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\cdots+{}^nC_ra^{n-r}b^r+\cdots+b^n$$

Look for the pattern, rather than just copying the formula. The power of $a$ starts at $n$ and drops by $1$ each term. The power of $b$ starts at $0$ and goes up by $1$ each term. In every term, the total degree stays at $n$, since $(n-r)+r=n$.

Expanding efficiently

To expand $(a+b)^n$, first identify the whole expression that acts as $a$, the whole expression that acts as $b$, and the value of $n$. If the second term is negative, keep it in brackets. For example, in $(2-3x)^3$, take $a=2$, $b=-3x$ and $n=3$, not just $b=3x$ with a minus sign floating around.

The general term is

For example, in $(1+2x)^4$, take $a=1$, $b=2x$ and $n=4$. The coefficients are $1,4,6,4,1$, so

$$(1+2x)^4=1+8x+24x^2+32x^3+16x^4$$

When a question asks for only the first few terms in ascending powers of $x$, don’t expand the whole thing. Use the early values $r=0,1,2,\ldots$ until you have enough terms. For instance, the first three terms of $(1-4x)^{10}$ come from $r=0$, $r=1$ and $r=2$:

$$(1-4x)^{10}=1+{}^{10}C_1(-4x)+{}^{10}C_2(-4x)^2+\cdots=1-40x+720x^2+\cdots$$

Finding a particular term

The binomial theorem is most useful when you skip expansion you don’t need. If you need a constant term, or the coefficient of a particular power of $x$, start with the general term and match the exponent.

In

$$\left(x+\frac{2}{x^3}\right)^8$$

the general term is

$${}^8C_r x^{8-r}\left(\frac{2}{x^3}\right)^r$$

The exponent of $x$ is $8-r-3r=8-4r$. A constant term has exponent $0$, so $8-4r=0$ and $r=2$. Then the constant term is

$${}^8C_2x^6\left(\frac{2}{x^3}\right)^2=112$$

That small exponent equation is often the neatest route. Expanding all nine terms wastes time and makes arithmetic errors more likely.

Pascal's triangle as coefficients

Pascal's triangle is a triangular array of numbers where each interior number is found by adding the two numbers directly above it. The rows give the binomial coefficients.

For row $n$, start counting rows from $0$. So row $4$ is

$$1,\ 4,\ 6,\ 4,\ 1$$

and these are the coefficients in the expansion of $(a+b)^4$:

$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$

The entries on the edges are $1$, since there is exactly one way to choose nothing and exactly one way to choose everything. The row symmetry helps too: choosing $r$ objects to include matches choosing $n-r$ objects to leave out.

Combinations and the formula for ${}^nC_r$

A combination is a selection of objects where order does not matter. The notation ${}^nC_r$ gives the number of combinations of $r$ objects chosen from $n$ distinct objects.

The formula is

For example,

$${}^6C_2=\frac{6!}{2!4!}=15$$

The subject guide expects you to find ${}^nC_r$ both by formula and by technology. On a non-calculator paper, use the factorial formula; it’s the safe method. On a calculator paper, use the combinations command or generate a table of values.

For the guide-style task of finding $r$ when

$${}^6C_r=20$$

a technology table for ${}^6C_r$, as $r$ runs from $0$ to $6$, shows that the value $20$ occurs at $r=3$. So $r=3$. Notice the symmetry in the row: ${}^6C_3$ is in the middle here, but in other rows the same value may appear twice because ${}^nC_r={}^nC_{n-r}$.

Why Pascal's triangle works

The connection between Pascal's triangle and combinations comes from counting the same thing in two cases. Suppose we want to choose $r$ people from a group of $n+1$ people. Pick one particular person, then split all possible selections into two non-overlapping cases:

• selections that include that person;
• selections that do not include that person.

If the chosen person is included, we still need $r-1$ more people from the remaining $n$, giving ${}^nC_{r-1}$ selections. If the chosen person is not included, all $r$ people must come from the remaining $n$, giving ${}^nC_r$ selections. Counting the same total this way gives

$${}^{n+1}C_r={}^nC_{r-1}+{}^nC_r$$

That is exactly the triangle-building rule: each interior entry is the sum of the two entries above it. It’s a neat mathematical generalization. A simple row pattern becomes a reliable algebraic tool for expanding any $(a+b)^n$ with $n\in\mathbb{N}$.

A note on naming and mathematical culture

The triangle is commonly called Pascal's triangle, but the pattern was known in several mathematical traditions before Pascal, including work associated with Yang Hui in China. That matters. Mathematics often gets presented through famous names, but good scholarship asks whether the name attached to an idea tells the whole story. Here it does not.

This also makes a useful TOK point: individuals can shape how mathematics is communicated and remembered, while the development of mathematical knowledge is usually wider than one person. The ethical issue is attribution: giving credit accurately is part of doing mathematics honestly.