This note shows how to expand a bracket raised to a positive whole-number power without multiplying it out, using binomial coefficients. It also shows how to pick out a single term, such as the coefficient of a chosen power of \(x\) or a constant term. These skills turn up in algebra, probability and approximation.
Key ideas
A binomial is an expression with two terms, such as \(2x+3\). Raising it to the power \(n\) gives an expansion with \(n+1\) terms. The powers of the first term go down from \(n\) to 0, while the powers of the second term go up from 0 to \(n\). In every term the two powers add to \(n\).
The numbers in front of each term are the binomial coefficients. You can find them in two ways.
- Pascal's triangle. Each row starts and ends with 1, and every other entry is the sum of the two entries above it. Row \(n\) (counting the top row as row 0) gives the coefficients for the power \(n\). Row 5 is 1, 5, 10, 10, 5, 1.
- The formula \(\binom{n}{r}\), read as "\(n\) choose \(r\)". It is best for large \(n\) because you do not need to build the whole triangle. Your GDC has an \({}^{n}C_{r}\) function, so you can find these values quickly.
The coefficients are symmetrical: \(\binom{n}{r}=\binom{n}{n-r}\). They also have simple end values, \(\binom{n}{0}=\binom{n}{n}=1\) and \(\binom{n}{1}=n\).
Treat each part of the bracket as one unit. In \((2x-3)^4\), the "first term" is \(2x\) and the "second term" is \(-3\). Put each in its own bracket when you substitute, so that powers and signs are applied to the whole term, including the number.
To find one particular term, do not expand everything. Write the general term, then choose \(r\) so that the power of \(x\) is the one you need.
Binomial expansions are also useful for estimating. If \(x\) is small, the first few terms of \((1+x)^n\) are very close to the full answer, because later terms contain tiny powers of \(x\).
Why the coefficients work
Think about multiplying out \((a+b)^3\) as \((a+b)(a+b)(a+b)\). To build a term, you choose either \(a\) or \(b\) from each of the three brackets. A term with one \(b\) and two \(a\)'s can be made in three ways, because the \(b\) can come from the first, second or third bracket. That is why the coefficient of \(a^2b\) is \(\binom{3}{1}=3\). In general, the coefficient of \(a^{n-r}b^r\) counts the ways of choosing which \(r\) brackets give a \(b\).
This also explains Pascal's triangle. To get the entry in row \(n\), you either keep the same number of \(b\)'s as in row \(n-1\) or add one more, so each entry is the sum of the two above it.
Useful checks
- If you set \(a=1\) and \(b=1\), the coefficients add up to \(2^n\). For row 4 this gives \(1+4+6+4+1=16=2^4\).
- If one bracket is \(1-x\), the signs of the terms alternate, so putting \(x=1\) gives a sum of 0 for \(n\ge 1\).
- For the middle of a long expansion, the coefficients are largest near \(r=n/2\), and they are symmetrical.
- A negative or fractional power such as \(x^{-3}\) can appear when a bracket contains \(\frac1x\). The powers still come from the same rule, \(a^{n-r}b^r\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(n!=n(n-1)(n-2)\cdots 2\times 1\) | Factorial. By definition \(0!=1\). |
| \(\dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!}\) | Number of ways to choose \(r\) items from \(n\). Also written \({}^{n}C_{r}\). Valid for whole numbers \(0\le r\le n\). |
| \((a+b)^n=\displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}b^{\,r}\) | The binomial theorem for \(n\) a positive integer. |
| \(T_{r+1}=\dbinom{n}{r}a^{\,n-r}b^{\,r}\) | General term. It is the \((r+1)\)th term, because the first term has \(r=0\). |
| \((1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\cdots\) | Useful special case for approximations when \(x\) is small. |
Worked examples
Example 1 Expand \((2x-3)^4\) in descending powers of \(x\).
Here \(a=2x\), \(b=-3\) and \(n=4\). The coefficients come from row 4 of Pascal's triangle: 1, 4, 6, 4, 1.
\((2x-3)^4=1(2x)^4+4(2x)^3(-3)+6(2x)^2(-3)^2+4(2x)(-3)^3+1(-3)^4\)
Evaluate each term: \((2x)^4=16x^4\); \(4\times 8x^3\times(-3)=-96x^3\); \(6\times 4x^2\times 9=216x^2\); \(4\times 2x\times(-27)=-216x\); \((-3)^4=81\).
Answer: \(16x^4-96x^3+216x^2-216x+81\)
Example 2 Find the coefficient of \(x^6\) and the constant term in the expansion of \(\left(3x^2+\dfrac{1}{x}\right)^6\).
The general term is \(\dbinom{6}{r}(3x^2)^{6-r}\left(\dfrac1x\right)^r=\dbinom{6}{r}3^{6-r}x^{12-2r}x^{-r}=\dbinom{6}{r}3^{6-r}x^{12-3r}\).
For \(x^6\): \(12-3r=6\), so \(r=2\). The term is \(\dbinom{6}{2}3^4x^6=15\times 81\,x^6=1215x^6\).
For the constant term: \(12-3r=0\), so \(r=4\). The term is \(\dbinom{6}{4}3^2=15\times 9=135\).
Answer: the coefficient of \(x^6\) is 1215 and the constant term is 135.
Example 3 Find the coefficient of \(x^3\) in the expansion of \((3-x)(1+2x)^7\).
The \(x^3\) term comes from two products: \(3\) times the \(x^3\) term of \((1+2x)^7\), and \(-x\) times the \(x^2\) term of \((1+2x)^7\).
The general term of \((1+2x)^7\) is \(\dbinom{7}{r}(2x)^r\).
\(x^3\) term: \(\dbinom73 2^3=35\times 8=280\), so \(x^3\) coefficient 280.
\(x^2\) term: \(\dbinom72 2^2=21\times 4=84\), so \(x^2\) coefficient 84.
Coefficient of \(x^3\) in the product: \(3\times 280+(-1)\times 84=840-84=756\).
Answer: 756
Common errors
- Forgetting that a negative second term changes signs. In \((2x-3)^4\) the terms alternate in sign, and \((-3)^3\) is negative.
- Raising only the letter to the power. In \((2x)^3\) the 2 is cubed too, giving \(8x^3\), not \(2x^3\).
- Using the wrong row of Pascal's triangle. The expansion of power \(n\) uses the row that starts 1, \(n\), so row 4 starts 1, 4.
- Mixing up \(r\) and the term number. The general term \(T_{r+1}\) is the \((r+1)\)th term, so the 4th term has \(r=3\).
- In a product of brackets, taking only one contribution to a coefficient. Check every way of making the required power.
Exam tip: When asked for a single term, write the general term first and simplify the power of \(x\) before solving for \(r\). It is quick, and it shows your method even if an arithmetic slip happens later.
Check yourself
- Find the coefficient of \(x^2\) in the expansion of \((1+x)^{10}\).
Answer
\(\dbinom{10}{2}=\dfrac{10\times 9}{2}=45\). - Find the coefficient of \(x^3\) in the expansion of \((2-x)^5\).
Answer
The term is \(\dbinom53 2^2(-x)^3=10\times 4\times(-1)x^3=-40x^3\). The coefficient is \(-40\). - Find the constant term in the expansion of \(\left(x+\dfrac1x\right)^6\).
Answer
The general term is \(\dbinom6r x^{6-r}x^{-r}=\dbinom6r x^{6-2r}\). Setting \(6-2r=0\) gives \(r=3\), so the constant term is \(\dbinom63=20\). - Use the first three terms of the expansion of \((1+0.01)^6\) to estimate \(1.01^6\).
Answer
\((1+x)^6\approx 1+6x+15x^2\). With \(x=0.01\): \(1+0.06+15\times 0.0001=1+0.06+0.0015=1.0615\).