- The binomial expansion writes \((a + b)^n\) as a sum of \(n + 1\) terms, \(\binom{n}{r} a^{n-r} b^r\), for \(r = 0, 1, \dots, n\).
- The coefficients \(\binom{n}{r}\), also written \({}^nC_r\), are the numbers in row \(n\) of Pascal’s triangle.
- To find one term, use the general term \(\binom{n}{r} a^{n-r} b^r\) and choose \(r\) to get the power you want. You never need the full expansion.
- In our case, \(1.02^{10} = (1 + 0.02)^{10} \approx 1 + 0.2 + 0.018 + 0.00096 = 1.21896\), so $5,000 saved at 2% a year grows to about $6,094.80 after 10 years (exact: $6,094.97).
- Exam point: keep signs and powers inside brackets, as in \((-2x)^3 = -8x^3\), not \(-2x^3\).
The case: what will a 2% savings account be worth?
Bilal puts $5,000 into a savings account that pays 2% interest a year, compounded once a year. He wants to know what it will be worth after 10 years, but he is sitting an exam where calculators are not allowed, and the question gives him only a pen. (The figures are illustrative.)
Each year the balance is multiplied by 1.02, so after 10 years it is
Working out \(1.02^{10}\) by repeated multiplication means ten rounds of long multiplication with growing numbers of decimal places. That is slow and error-prone. The binomial expansion gives a much faster route: write \(1.02\) as \(1 + 0.02\), expand, and keep only the first few terms. Because 0.02 is small, its higher powers are tiny, and three or four terms already give an answer accurate to the nearest few cents per hundred dollars.
This post shows how the expansion works, how to find any single term quickly, and how to judge how many terms an approximation needs. (This post is about the binomial theorem in algebra. The binomial distribution in statistics uses the same coefficients, and we link to it at the end.)
The concept: expanding a bracket raised to a power
Expand \((a + b)^2\) and you get \(a^2 + 2ab + b^2\). Expand \((a + b)^3\) and you get \(a^3 + 3a^2b + 3ab^2 + b^3\). Two patterns stand out. First, the powers of \(a\) fall by one each term while the powers of \(b\) rise by one, and in every term the two powers add up to \(n\). Second, the coefficients follow a pattern: 1, 2, 1, then 1, 3, 3, 1.
Those coefficients are the rows of Pascal’s triangle. Each row starts and ends with 1, and every other number is the sum of the two numbers above it:
| \(n\) | Coefficients |
|---|---|
| 0 | 1 |
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
The reason is counting. When you multiply out \(n\) brackets of \((a + b)\), each term is made by picking either \(a\) or \(b\) from every bracket. The number of ways to pick \(b\) from exactly \(r\) of the \(n\) brackets is \(\binom{n}{r}\), “\(n\) choose \(r\)”. So that is the coefficient of \(a^{n-r}b^r\).[3]
Here \(n\) is the power, \(a\) and \(b\) are the two parts of the bracket, and \(r\) counts how many factors of \(b\) a term contains. The term with \(r\) factors of \(b\) is the \((r + 1)\)th term, so the general term is \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\). For the special case \(a = 1\), the formula becomes \((1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots\), which is the form we use for approximations.[1]
Mathematics Made Simple explains binomial expansion and Pascal’s triangle with worked examples and graphs.Get the textbook · $12
For small \(n\), reading a row of Pascal’s triangle is quickest. For larger \(n\), or when you need only one term, the formula for \(\binom{n}{r}\) is better. A useful shortcut: \(\binom{n}{r} = \frac{n(n-1)\cdots(n-r+1)}{r!}\), so \(\binom{10}{3} = \frac{10 \times 9 \times 8}{6} = 120\). The coefficients are also symmetric: \(\binom{n}{r} = \binom{n}{n-r}\).
Why this method fits the case
The savings problem has the exact shape \((1 + x)^n\) with a small \(x\). In that form the terms shrink fast. With \(x = 0.02\) and \(n = 10\), the first four terms are \(1\), \(0.2\), \(0.018\) and \(0.00096\). The fifth term is \(\binom{10}{4}(0.02)^4 = 210 \times 0.00000016 = 0.0000336\), already tiny: on $5,000 it is worth only about 17 cents. So a few lines of hand arithmetic give a very good answer.
The method also shows where the answer comes from. The term \(nx = 0.2\) is simple interest: ten years at 2%. The next term, \(0.018\), is the “interest on interest” that compounding adds. Seeing the pieces separately helps students understand why compound growth beats simple growth, and by how much.[4]
When is it not the right tool? The approximation is good only when \(nx\) is fairly small. If the rate is large or the number of years is big, more and more terms are needed, and a calculator or logarithms are better. For \(1.02^{30}\), three terms give 1.774 while the true value is about 1.811, an error of 2%. The expansion we use here is also only for whole-number powers. For powers such as \(\tfrac{1}{2}\) or \(-1\), the series never stops and needs \(|x| < 1\); that extension belongs to a later topic.[3]
Setting it up
Here are the quantities in the case.
| Quantity | Symbol | Value |
|---|---|---|
| Amount deposited | \(P\) | $5,000 |
| Annual interest rate | \(x\) | 2% = 0.02 |
| Number of years | \(n\) | 10 |
| Growth factor | \((1 + x)^n\) | \(1.02^{10}\) |
| Final balance | \(P(1 + x)^n\) | to be found |
We need \(\binom{10}{0} = 1\), \(\binom{10}{1} = 10\), \(\binom{10}{2} = 45\) and \(\binom{10}{3} = 120\).
Solving it
We start with a full expansion, then find a single term, then solve the savings case.
Expand \((2 + 3x)^4\) fully.
\((2 + 3x)^4 = 16 + 96x + 216x^2 + 216x^3 + 81x^4\). Check: put \(x = 1\); \(5^4 = 625\) and \(16 + 96 + 216 + 216 + 81 = 625\).
Find the coefficient of \(x^3\) in \((1 – 2x)^8\), and the term independent of \(x\) in \(\left(x + \dfrac{2}{x}\right)^6\).
The coefficient of \(x^3\) is −448, and the constant term is 160.
Use the first four terms of the expansion of \((1 + x)^{10}\) to estimate \(1.02^{10}\), and hence the value of $5,000 after 10 years at 2%.
After 10 years Bilal’s $5,000 is worth about $6,094.80, within 17 cents of the exact figure.
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What the answer tells us
Each term in the expansion has a meaning. The 1 is the original deposit. The \(0.2\) is ten years of simple interest at 2%: $1,000 on $5,000. The \(0.018\) is the first layer of interest on interest, worth $90. The \(0.00096\) adds another $4.80. So compounding adds $94.97 in total over simple interest, and nearly all of it comes from the \(x^2\) term.
Reasonableness check. Simple interest alone gives a factor of 1.2, and compounding can only add to that, so the answer must be a little above 1.2. It is: 1.219. Also, every term we dropped is positive, so the estimate must be slightly below the true value, which it is.
Sensitivity. How many terms do we need? The table below compares the exact growth factor with the two-term (simple interest) and three-term approximations.
| Years \(n\) | Exact \(1.02^n\) | 2 terms: \(1 + 0.02n\) | 3 terms |
|---|---|---|---|
| 10 | 1.2190 | 1.2000 | 1.2180 |
| 20 | 1.4859 | 1.4000 | 1.4760 |
| 30 | 1.8114 | 1.6000 | 1.7740 |
Over 10 years, three terms are within 0.1%. Over 30 years, three terms are about 2% low, and four terms (1.8065) are needed to get within 0.3%. As \(nx\) grows, later terms stop being small, so the approximation needs more of them. The chart shows the same story.
The Mathematics Made Simple textbook builds the binomial theorem from Pascal’s triangle and links it to sequences and series. For practice on full expansions, single terms and approximations, the Mathematics Made Simple Practice Questions book has graded sets with every line worked.
Master sequences, series and the binomial theorem with both Mathematics Made Simple books for $16.Both books · $16
Practice questions
Work each question by hand first. Then open the worked answer to check every coefficient.
EasyQ1. Write down row 5 of Pascal’s triangle and use it to expand \((1 + x)^5\). Show worked answer
\((1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5\).
Check: at \(x = 1\), both sides equal 32.
EasyQ2. Find the coefficient of \(x^2\) in the expansion of \((3 + x)^5\). Show worked answer
\(\binom{5}{2} \times 3^3 = 10 \times 27 = 270\).
The coefficient is 270.
MediumQ3. Find the term independent of \(x\) in \(\left(2x – \dfrac{1}{x}\right)^6\). Show worked answer
Power 0: \(6 – 2r = 0\), so \(r = 3\).
\(\binom{6}{3} \times 2^3 \times (-1)^3 = 20 \times 8 \times (-1) = -160\).
MediumQ4. The first three terms of \((1 + ax)^n\) are \(1 + 12x + 60x^2\). Find \(n\) and \(a\). Show worked answer
From the first, \(a = \frac{12}{n}\). Substitute: \(\frac{n(n-1)}{2} \times \frac{144}{n^2} = 60\), so \(\frac{72(n-1)}{n} = 60\).
\(72n – 72 = 60n\), giving \(n = 6\) and \(a = 2\).
Check: \(\binom{6}{2} \times 4 = 60\).
ExamQ5. (a) Find the first three terms, in ascending powers of \(x\), of the expansion of \((2 – x)^6\). [3] (b) Hence find the coefficient of \(x^2\) in \((1 + 3x)(2 – x)^6\). [2] (c) Use your answer to (a) to estimate \(1.99^6\). [2] Show worked answer
(b) \(1 \times 240 + 3 \times (-192)\) [M1] \(= 240 – 576 = -336\) [A1].
(c) \(1.99 = 2 – 0.01\), so put \(x = 0.01\) [M1]: \(64 – 1.92 + 0.024 = 62.104\) [A1]. (Exact: 62.1038 to 4 d.p.)
Frequently asked questions
What is the difference between nCr and Pascal’s triangle?
They give the same numbers. Row \(n\) of Pascal’s triangle lists \(\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}\). The triangle is quicker for small \(n\); the formula is quicker when \(n\) is large or you need only one coefficient.
How many terms are in the expansion of (a + b)ⁿ?
There are \(n + 1\) terms, because \(r\) runs from 0 to \(n\). So \((a + b)^{10}\) has 11 terms. If \(a\) and \(b\) contain the same variable, some terms may combine, but before simplifying there are \(n + 1\).
Is the binomial theorem the same as the binomial distribution?
They are linked but not the same. The binomial distribution in statistics gives probabilities \(\binom{n}{r}p^r(1-p)^{n-r}\), which are exactly the terms of \((p + (1 – p))^n\). The theorem is the algebra; the distribution is one application of it.
How do I know how many terms to use in an approximation?
Keep adding terms until the next one is smaller than the accuracy you need. With \(x = 0.02\) and \(n = 10\), the fourth term is about 0.001 and the fifth is about 0.00003, so four terms give the growth factor to about four decimal places.
References
- Abramson, J. et al. (2021) Algebra and Trigonometry 2e. Houston: OpenStax. https://openstax.org/details/books/algebra-and-trigonometry-2e
- Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
- International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide. Geneva: IBO.
- Sydsæter, K., Hammond, P., Strøm, A. and Carvajal, A. (2016) Essential Mathematics for Economic Analysis, 5th edn. Harlow: Pearson.
