Enter A, B and n of the expression (Ax+B)ⁿ to expand. The formula below is linked to the input fields, so you can also edit the formula directly. Use "Find" to choose between expanding every term and finding just one term.
Enter numbers using digits. Decimals, negative numbers and fractions such as 3/4 are fine. The exponent n is a whole number from 0 to 12.
Result and figure
Enter the coefficients and the exponent in the fields on the left and press "Calculate". The expanded expression and Pascal's triangle will appear here.
Table of Contents
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What you can do on this page
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What is this calculation used for?
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How to Use
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Formulas and figures
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Symbols and terms
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Good to know before you start
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How to calculate it in Excel
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How to calculate it in Google Sheets
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How to calculate it in Python
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How to write it in LaTeX and other math languages (copy and paste)
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How to have ChatGPT do the calculation
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DataChef Features
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Related Features
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NumberChef Calculators List
What you can do on this page
- Expand \((Ax+B)^{n}\) (\(n\) from 0 to 12) with the binomial theorem and see the result in descending powers of \(x\) right away
- If you only need the coefficient of \(x^{3}\), give the power of \(x\) (or the position of the term) and find just that one term
- When you find one term, the value of the binomial coefficient \({}_n\mathrm{C}_k\) and the steps using \(\dfrac{n!}{k!\,(n-k)!}\) are shown too
- Pascal's triangle is drawn down to row \(n\), with the row you used highlighted
- \(A\) and \(B\) can be decimals, negative numbers or fractions such as 3/4. Coefficients are calculated exactly, as simplified fractions
- A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page
This page handles the form \((Ax+B)^{n}\) (\(n\) is a whole number from 0 to 12). It does not handle fractional or negative exponents (the general binomial theorem) or expressions with three or more terms (the multinomial theorem).
What is this calculation used for?
The probability of k successes in n tries (quality control, polls, clinical trials)
Repeat a trial \(n\) times, where the chance \(p\) is the same every time and one result does not affect another. The probability that something happens exactly \(k\) times is \({}_n\mathrm{C}_k p^{k}(1-p)^{n-k}\). This is exactly one term of \((p + (1-p))^n\) expanded with the binomial theorem, and this kind of probability distribution is called the binomial distribution.
For example, if a production line always has a 3% defect rate and you sample 10 items, the chance of no defects at all is the \(k = 0\) term, \(0.97^{10} \approx 0.737\), or about 74%. Put the other way, checking 10 items finds a defect about 26% of the time. The same idea underlies deciding how many items to inspect, estimating the margin of error of a poll, and judging whether a new drug works.
Predicting traits in offspring (biology and agriculture)
When both parents have the gene pair \(\mathrm{Aa}\), their children's genotypes appear in the ratio \(\mathrm{AA} : \mathrm{Aa} : \mathrm{aa} = 1 : 2 : 1\). The 1, 2, 1 comes from each parent passing on \(\mathrm{A}\) or \(\mathrm{a}\) with probability \(\dfrac{1}{2}\) each: \(\left(\dfrac{1}{2}\mathrm{A} + \dfrac{1}{2}\mathrm{a}\right)^{2} = \dfrac{1}{4}\left(\mathrm{AA} + 2\,\mathrm{Aa} + \mathrm{aa}\right)\). These are the binomial coefficients, the same numbers as row \(n = 2\) of Pascal's triangle.
With more traits, or when you break it down by the number of children, it gets too hard to track by hand. The binomial theorem handles questions like "the chance that exactly 2 of 4 siblings have a trait" in one step. Plant and animal breeders and genetic counselors estimate expected ratios this way.
Quick estimates of compound interest and growth (money and economics)
Money growing at an annual rate \(r\) for \(n\) years is multiplied by \((1+r)^{n}\). Expanding this with the binomial theorem gives \(1 + nr + {}_n\mathrm{C}_2 r^{2} + \cdots\). The first part, \(1 + nr\), is simple interest, and the rest is "interest on interest".
For example, at 1% a year for 10 years, \(1.01^{10} = 1.10462\cdots\), but the first three terms alone, \(1 + 0.1 + 0.0045\), give 1.1045, almost the same value. The later terms get smaller and smaller, so this is handy for a rough mental estimate. Interest rates, inflation rates and population growth rates, anything that grows by the same percentage each year, all take this form.
Estimating the effect of small changes (engineering, physics, measurement)
In design and measurement, you often need to know quickly how much a result changes when something changes slightly. Expand with the binomial theorem and drop the small squared and higher terms, and you get the approximation \((1+x)^{n} \approx 1 + nx\).
For example, if each edge of a cube gets 1% longer, the volume becomes \((1+0.01)^{3} = 1.030301\) times as large, an increase of about 3%. The approximation gives \(1 + 3 \times 0.01 = 1.03\), almost the same answer. The rule of thumb "a 1% error in length means about a 3% error in volume" is used directly when setting part tolerances and estimating measurement errors.
Counting all the possible choices (combinations and counting)
When you freely decide "yes or no" for each of \(n\) options, the total number of combinations is \(2^{n}\). This is the binomial theorem with \(a = b = 1\), \(\sum_{k=0}^{n} {}_n\mathrm{C}_k = 2^{n}\), and it is also the sum of each row of Pascal's triangle.
For example, a pizza with 5 available toppings can be ordered \(2^{5} = 32\) ways, including a plain cheese pizza with no toppings. The row also gives the breakdown: exactly 2 toppings can be ordered in \({}_5\mathrm{C}_2 = 10\) ways. This idea is used whenever you count choices, such as menu combinations, survey answer patterns and software test cases.
Formulas and figures
The binomial theorem
Figure
Standard notation
(the usual math form)
\((a+b)^{n}\)
\(=\)
\(\displaystyle\sum_{k=0}^{n}\)
\({}_{n}\mathrm{C}_{k}\)
\(a^{n-k}\)
\(b^{k}\)
In words
(symbols replaced with words)
⑤
\((a+b)^n\): \(n\) copies of \((a+b)\) multiplied together
\(=\)
④
add them all up, with \(k\), the number of \(b\)s chosen, going from 0 to \(n\)
①
\({}_n\mathrm{C}_k\): ways to choose \(k\) of \(n\)
②
\(a^{n-k}\): \((n-k)\) copies of \(a\) multiplied
③
\(b^{k}\): \(k\) copies of \(b\) multiplied
The formula in words
①
Multiply the
\({}_n\mathrm{C}_k\): ways to choose \(k\) of \(n\)
②
by
\(a^{n-k}\): \((n-k)\) copies of \(a\) multiplied
③
and by
\(b^{k}\): \(k\) copies of \(b\) multiplied
④
then
add them all up, with \(k\), the number of \(b\)s chosen, going from 0 to \(n\)
⑤
and you get
\((a+b)^n\): \(n\) copies of \((a+b)\) multiplied together
Quick example
Expanding \((a+b)^{3}\) (the coefficients 1, 3, 3, 1 are the values of \({}_3\mathrm{C}_0\) to \({}_3\mathrm{C}_3\)) gives
3 copies of \((a+b)\) multiplied together
\(=\)
add them all up, with the number of \(b\)s \(k\) going from 0 to 3
ways to choose \(k\) of 3
\((3-k)\) copies of \(a\) multiplied
\(k\) copies of \(b\) multiplied
\((a+b)^{3} = {}_{3}\mathrm{C}_{0}a^{3} + {}_{3}\mathrm{C}_{1}a^{2}b + {}_{3}\mathrm{C}_{2}ab^{2} + {}_{3}\mathrm{C}_{3}b^{3}\)
\(= a^{3} + 3a^{2}b + 3ab^{2} + b^{3}\)
Key idea
\((a+b)^n\) means "write the parentheses \((a+b)\) \(n\) times and multiply them". To expand, you pick either \(a\) or \(b\) from each set of parentheses, multiply your picks together, and add up the results of every possible way of picking.
The number of ways to choose which \(k\) of the \(n\) parentheses give you a \(b\) is \({}_n\mathrm{C}_k\). So the same kind of term \(a^{n-k}b^{k}\) appears \({}_n\mathrm{C}_k\) times, and when you combine them, the coefficient is \({}_n\mathrm{C}_k\). That is all the binomial theorem says. (In many US textbooks, \({}_n\mathrm{C}_k\) is written \(\binom{n}{k}\) and read "n choose k".)
The figure above shows \((a+b)^2\) as the area of a square. There are two \(ab\) rectangles, which gives \(2ab\), so you can see the coefficients 1, 2, 1 with your own eyes.
The general term (the \((k+1)\)th term)
Figure
Standard notation
(the usual math form)
\(T_{k+1}\)
\(=\)
\({}_{n}\mathrm{C}_{k}\)
\(a^{n-k}\)
\(b^{k}\)
In words
(symbols replaced with words)
④
\(T_{k+1}\): the \((k+1)\)th term of the expansion
\(=\)
①
\({}_n\mathrm{C}_k\): ways to choose \(k\) of \(n\)
②
\(a^{n-k}\): \((n-k)\) copies of \(a\) multiplied
③
\(b^{k}\): \(k\) copies of \(b\) multiplied
The formula in words
①
Multiply the
\({}_n\mathrm{C}_k\): ways to choose \(k\) of \(n\)
②
by
\(a^{n-k}\): \((n-k)\) copies of \(a\) multiplied
③
and by
\(b^{k}\): \(k\) copies of \(b\) multiplied
④
and you get
\(T_{k+1}\): the \((k+1)\)th term of the expansion
Quick example
The 3rd term (\(k = 2\)) of \((2x+3)^{5}\), with \(a = 2x\) and \(b = 3\), is
3rd term \(T_{3}\)
\(=\)
ways to choose 2 of 5
3 copies of \(2x\) multiplied
2 copies of 3 multiplied
\(T_{3} = {}_{5}\mathrm{C}_{2}(2x)^{3}\cdot 3^{2} = 10 \times 8x^{3} \times 9 = 720x^{3}\)
Key idea
The most common mistake with the binomial theorem is that the term number and \(k\) are off by one. The term with \(k = 0\) is the 1st term, so in the \((k+1)\)th term, \(k\) is "the term number minus 1". For the 3rd term, \(k = 2\).
When a problem asks for "the coefficient of \(x^{3}\)", find \(k\) from the power of \(x\), not from the term number. In \((Ax+B)^n\), the power of \(x\) is \(n-k\), so solve \(n-k = 3\) to get \(k = n-3\).
When \(a\) or \(b\) is a group such as \(2x\), raise the whole group to the power, as in \((2x)^{3} = 8x^{3}\). To avoid writing \(2x^{3}\) by mistake, always put parentheses around it before you calculate.
Calculating a binomial coefficient (the number of combinations)
Standard notation
(the usual math form)
\({}_{n}\mathrm{C}_{k}\)
\(=\)
\(\dfrac{n!}{k!\,(n-k)!}\)
In words
(symbols replaced with words)
②
\({}_n\mathrm{C}_k\): ways to choose \(k\) of \(n\)
\(=\)
①
\(\dfrac{\text{ways to line up all }n\text{ items }n!}{\text{orders of the }k\text{ chosen }k!\ \times\ \text{orders of the }(n-k)\text{ left }(n-k)!}\)
The formula in words
①
The
\(\dfrac{\text{ways to line up all }n\text{ items }n!}{\text{orders of the }k\text{ chosen }k!\ \times\ \text{orders of the }(n-k)\text{ left }(n-k)!}\)
②
is the
\({}_n\mathrm{C}_k\): ways to choose \(k\) of \(n\)
Quick example
The number of ways to choose 2 of 5, \({}_5\mathrm{C}_2\), is
ways to choose 2 of 5
\(=\)
\(\dfrac{\text{ways to line up all 5 }5!}{\text{orders of the 2 chosen }2!\ \times\ \text{orders of the 3 left }3!}\)
\({}_{5}\mathrm{C}_{2} = \dfrac{5!}{2!\,(5-2)!} = \dfrac{120}{2 \times 6} = 10\)
\({}_{5}\mathrm{C}_{2} = \dfrac{5 \times 4}{2 \times 1} = 10\)
Key idea
A factorial is the product of every whole number from 1 up to that number (\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)). By definition \(0! = 1\), so \({}_n\mathrm{C}_0 = {}_n\mathrm{C}_n = 1\).
Why divide? Think about lining things up. Put all \(n\) items in a row, and say the first \(k\) are "chosen" and the other \((n-k)\) are "not chosen". Then each of the \(n!\) orders gives a choice. But reordering the \(k\) chosen items (\(k!\) ways) or the \((n-k)\) items left (\((n-k)!\) ways) does not change who was chosen. Each choice is counted \(k!\,(n-k)!\) times, so dividing by that leaves just the number of choices.
By hand, the second form above is faster: the top counts down from \(n\) for \(k\) numbers, and the bottom is \(k!\). For \({}_5\mathrm{C}_2\), that is \(\dfrac{5 \times 4}{2 \times 1} = 10\).
Binomial coefficients are symmetric: \({}_n\mathrm{C}_k = {}_n\mathrm{C}_{n-k}\) (choosing \(k\) items is the same as choosing the \((n-k)\) items to leave out). For example, \({}_7\mathrm{C}_5\) is quicker as \({}_7\mathrm{C}_2 = 21\). This is also why the coefficients of an expansion read the same from both ends.
Pascal's triangle (the sum of two neighbors)
Figure
Standard notation
(the usual math form)
\({}_{n-1}\mathrm{C}_{k-1}\)
\(+\)
\({}_{n-1}\mathrm{C}_{k}\)
\(=\)
\({}_{n}\mathrm{C}_{k}\)
In words
(symbols replaced with words)
①
\({}_{n-1}\mathrm{C}_{k-1}\): upper-left number in the row above
\(+\)
②
\({}_{n-1}\mathrm{C}_{k}\): upper-right number in the row above
\(=\)
③
\({}_n\mathrm{C}_k\): the number just below them
The formula in words
①
Add the
\({}_{n-1}\mathrm{C}_{k-1}\): upper-left number in the row above
②
and the
\({}_{n-1}\mathrm{C}_{k}\): upper-right number in the row above
③
and you get
\({}_n\mathrm{C}_k\): the number just below them
Quick example
Adding the 3 and 3 in row \(n = 3\) gives the 6 in row \(n = 4\)
upper left \({}_3\mathrm{C}_1 = 3\)
\(+\)
upper right \({}_3\mathrm{C}_2 = 3\)
\(=\)
just below \({}_4\mathrm{C}_2 = 6\)
\({}_{3}\mathrm{C}_{1} + {}_{3}\mathrm{C}_{2} = 3 + 3 = 6 = {}_{4}\mathrm{C}_{2}\)
Key idea
Pascal's triangle is the binomial coefficients arranged in a triangle, row by row from the top. Both ends of every row are 1, and each number inside is just the sum of the two numbers directly above it (upper left and upper right). You can build the coefficients with addition alone, without any factorials.
The choosing idea also explains why this addition works. To choose \(k\) of \(n\) items, focus on one particular item. Either you choose it (then choose \((k-1)\) more from the other \((n-1)\)), or you do not (then choose all \(k\) from the other \((n-1)\)). The two cases are exactly the two numbers above.
The numbers in each row add up to \(2^n\) (row \(n = 4\): \(1+4+6+4+1 = 16 = 2^4\)). This is the binomial theorem with \(a = b = 1\).
The binomial theorem says \((a+b)^n = \sum_{k=0}^{n} {}_n\mathrm{C}_k a^{n-k} b^{k}\): in the expansion, the coefficient of \(a^{n-k}b^{k}\) is \({}_n\mathrm{C}_k\). To find just one term, use the general term \({}_n\mathrm{C}_k a^{n-k} b^{k}\) and choose \(k\) from the power of \(x\). The coefficients match row \(n\) of Pascal's triangle.
Symbols and terms
Symbols
| \({}_n\mathrm{C}_k\) | n C k, or n choose k | The number of ways to choose \(k\) things from \(n\) (the number of combinations). In the binomial theorem it becomes a coefficient of the expansion, so it is also called a binomial coefficient. \(\mathrm{C}\) is the first letter of "combination". It is also written \(\binom{n}{k}\), nCk or C(n, k). |
| \(n!\) | n factorial | The product of every whole number from 1 to \(n\) (\(4! = 4 \times 3 \times 2 \times 1 = 24\)). It looks like an exclamation mark, but it is the math symbol for "factorial". By definition, \(0! = 1\). |
| \(a,\ b\) | a, b | The two terms inside the parentheses in the binomial theorem. "Binomial" means "two terms", and these are the two. By custom, letters from the start of the alphabet, \(a,\ b,\ c\), stand for fixed values. |
| \(n\) | n | How many copies of the same parentheses are multiplied (the exponent). The letter \(n\), from "number", is often used for a count. On this page it is a whole number from 0 to 12. |
| \(k\) | k | A counter for "how many \(b\)s were chosen". It runs from 0 to \(n\). The term with \(k = 0\) is the 1st term, which is why we say "the \((k+1)\)th term". The letter \(k\) is a common choice for a whole-number counter. |
| \(\displaystyle\sum\) | sigma | The symbol for "add them all up". It is the capital Greek letter sigma, which matches S for "sum". \(\sum_{k=0}^{n}\) means "add up the expression after it, with \(k\) going from 0 to \(n\)". |
| \(T_{k+1}\) | T sub k plus 1 | The term in position \(k+1\) from the start when the expansion is written in descending powers. \(T\) is the first letter of "term". It is called the general term. |
| \(A,\ B\) | capital A, capital B | The letters used in this calculator's input fields. In \((Ax+B)^n\), \(A\) is the coefficient of \(x\) and \(B\) is the constant term without \(x\). In the binomial theorem, \(a = Ax\) and \(b = B\). |
| \(x\) | x | A letter whose value is not fixed (a variable). This calculator expands expressions of the form \((Ax+B)^n\), so the answer is a polynomial in \(x\). |
Terms
| binomial theorem | The theorem that the coefficients in the expansion of \((a+b)^n\) are the numbers of combinations \({}_n\mathrm{C}_k\). In the US it is usually taught in Algebra 2 or Precalculus. |
| binomial coefficient | The coefficient \({}_n\mathrm{C}_k\) that appears in the binomial theorem. It is the same as the number of combinations, but this name is used when you think of it as a coefficient of an expansion. |
| combination | A way of choosing some items from a group without caring about order. The number of ways to choose \(k\) of \(n\) is written \({}_n\mathrm{C}_k\). Unlike a permutation, where order matters, the same set of items counts as one way. |
| factorial | The product of every whole number from 1 up to that number. \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). It is used to calculate the number of combinations. |
| general term | A single formula that describes every term of an expansion by its position. In the binomial theorem, the \((k+1)\)th term \({}_n\mathrm{C}_k a^{n-k} b^{k}\) is the general term. To find one term, put values into it. |
| expansion | Multiplying out an expression with parentheses to write it as a sum of terms without parentheses. It is the exact opposite of factoring. |
| descending powers | Writing the terms of an expression from the highest power of the variable down, as in \(x^3 + 2x^2 - x + 5\). Answers are usually written in this order (also called standard form). |
| Pascal's triangle | The binomial coefficients arranged row by row in a triangle. Both ends are 1, and each number inside is the sum of the two numbers above it. It is named after the French mathematician Blaise Pascal, but it was known centuries earlier in Persia, India and China, where it is called Yang Hui's triangle. |
| polynomial | An expression that adds up terms made of a coefficient times a power of \(x\), such as \(2x^3 - x + 5\). The result of an expansion with the binomial theorem is a polynomial too. |
| term | Each piece of an expression separated by plus and minus signs. \(x^3 - 2x + 3\) has three terms: \(x^3\), \(-2x\) and \(3\). |
| coefficient | The number in front of a variable. In \(3x^2\), the coefficient is 3. When no number is written, as in \(x^2\), the coefficient is 1. |
| exponent | The small 3 written at the upper right of \(x^3\). It shows how many times the same thing is multiplied (\(x^3 = x \times x \times x\)). Also called the power. |
| constant term | A term with no \(x\), only a number, like the 5 in \(x^2 + 3x + 5\). In this calculator, \(B\) inside the parentheses and the last term of the expansion are constant terms. |
| binomial distribution | The probability distribution of the number of successes \(k\) when the same trial is repeated \(n\) times. Its probabilities \({}_n\mathrm{C}_k p^{k}(1-p)^{n-k}\) are exactly the terms of \((p + (1-p))^n\) expanded with the binomial theorem, which is where the name "binomial" comes from. |
| general binomial theorem | The binomial theorem extended to fractional and negative exponents. The result is an expression with infinitely many terms (an infinite series). This calculator handles only whole-number exponents of 0 or more. |
Good to know before you start
Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back over these topics is the quickest way forward.
| Multiplying out expressions (Grade 8 to Algebra 1) |
|
| Powers and the laws of exponents (Grade 8 to Algebra 1) |
|
| Working with negative numbers (Grades 6–7) |
|
| Combinations and factorials (Algebra 2 or Precalculus) |
|
| Sigma notation (Precalculus) |
|
| Working with fractions (Grades 5–6) |
|
How to calculate it in Excel
Copy the whole table below and paste it into cell A1 in Excel. It works as is.
Table to find each coefficient of the expansion with the binomial theorem
| Coefficient of x, A | 2 |
| Constant term B | 3 |
| Exponent n | 5 |
| Coefficient for k=0 (x to the n−0) | =IFERROR(COMBIN($B$3,0)*$B$1^($B$3-0)*$B$2^0,0) |
| Coefficient for k=1 (x to the n−1) | =IFERROR(COMBIN($B$3,1)*$B$1^($B$3-1)*$B$2^1,0) |
| Coefficient for k=2 (x to the n−2) | =IFERROR(COMBIN($B$3,2)*$B$1^($B$3-2)*$B$2^2,0) |
| Coefficient for k=3 (x to the n−3) | =IFERROR(COMBIN($B$3,3)*$B$1^($B$3-3)*$B$2^3,0) |
| Coefficient for k=4 (x to the n−4) | =IFERROR(COMBIN($B$3,4)*$B$1^($B$3-4)*$B$2^4,0) |
| Coefficient for k=5 (x to the n−5) | =IFERROR(COMBIN($B$3,5)*$B$1^($B$3-5)*$B$2^5,0) |
| Coefficient for k=6 (x to the n−6) | =IFERROR(COMBIN($B$3,6)*$B$1^($B$3-6)*$B$2^6,0) |
Table to find the coefficient of the general term ((k+1)th term)
| Coefficient of x, A | 2 |
| Constant term B | 3 |
| Exponent n | 5 |
| Term number (from the start) | 3 |
| k (term number minus 1) | =B4-1 |
| Power of x (n−k) | =B3-B5 |
| Binomial coefficient nCk | =COMBIN(B3,B5) |
| Coefficient of the term | =B7*B1^B6*B2^B5 |
Table to find a binomial coefficient (number of combinations)
| n | 5 |
| k | 2 |
| Binomial coefficient nCk | =COMBIN(B1,B2) |
| Check with factorials | =FACT(B1)/(FACT(B2)*FACT(B1-B2)) |
Table to build Pascal's triangle
| Row n = 0 | 1 | |||||
| Row n = 1 | 1 | 1 | ||||
| Row n = 2 | 1 | =B2+C2 | 1 | |||
| Row n = 3 | 1 | =B3+C3 | =C3+D3 | 1 | ||
| Row n = 4 | 1 | =B4+C4 | =C4+D4 | =D4+E4 | 1 | |
| Row n = 5 | 1 | =B5+C5 | =C5+D5 | =D5+E5 | =E5+F5 | 1 |
After pasting, the upper rows (A, B, n and so on) are your inputs, and the lower rows are calculated automatically.
COMBIN(n, k) is the binomial coefficient nCk, FACT(n) is n factorial, and "^" is a power.
The first table expands (2x+3)⁵. From k=0 down, it shows 32, 240, 720, 1080, 810, 243 (the row for k is the coefficient of x to the power n−k). Change A, B and n, and you get the coefficients of your own expression. When n is less than 6, rows for terms that do not exist show 0 (IFERROR turns them into 0). When n is 7 or more, copy the last row down and change the 6 in the formula to 7, 8 and so on.
The second table finds only the 3rd term: k is 2, the power of x is 3, the binomial coefficient is 10, and the coefficient of the term is 720.
The third table is the number of ways to choose 2 of 5, and both formulas give 10.
In the fourth table, only the 1s at both ends are typed in. Each number inside is a formula that adds the two cells just above it. Row n = 5 shows 1, 5, 10, 10, 5, 1.
COMBIN(n, k) is the binomial coefficient nCk, FACT(n) is n factorial, and "^" is a power.
The first table expands (2x+3)⁵. From k=0 down, it shows 32, 240, 720, 1080, 810, 243 (the row for k is the coefficient of x to the power n−k). Change A, B and n, and you get the coefficients of your own expression. When n is less than 6, rows for terms that do not exist show 0 (IFERROR turns them into 0). When n is 7 or more, copy the last row down and change the 6 in the formula to 7, 8 and so on.
The second table finds only the 3rd term: k is 2, the power of x is 3, the binomial coefficient is 10, and the coefficient of the term is 720.
The third table is the number of ways to choose 2 of 5, and both formulas give 10.
In the fourth table, only the 1s at both ends are typed in. Each number inside is a formula that adds the two cells just above it. Row n = 5 shows 1, 5, 10, 10, 5, 1.
How to calculate it in Google Sheets
Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.
Table to find each coefficient of the expansion with the binomial theorem
| Coefficient of x, A | 2 |
| Constant term B | 3 |
| Exponent n | 5 |
| Coefficient for k=0 (x to the n−0) | =IFERROR(COMBIN($B$3,0)*$B$1^($B$3-0)*$B$2^0,0) |
| Coefficient for k=1 (x to the n−1) | =IFERROR(COMBIN($B$3,1)*$B$1^($B$3-1)*$B$2^1,0) |
| Coefficient for k=2 (x to the n−2) | =IFERROR(COMBIN($B$3,2)*$B$1^($B$3-2)*$B$2^2,0) |
| Coefficient for k=3 (x to the n−3) | =IFERROR(COMBIN($B$3,3)*$B$1^($B$3-3)*$B$2^3,0) |
| Coefficient for k=4 (x to the n−4) | =IFERROR(COMBIN($B$3,4)*$B$1^($B$3-4)*$B$2^4,0) |
| Coefficient for k=5 (x to the n−5) | =IFERROR(COMBIN($B$3,5)*$B$1^($B$3-5)*$B$2^5,0) |
| Coefficient for k=6 (x to the n−6) | =IFERROR(COMBIN($B$3,6)*$B$1^($B$3-6)*$B$2^6,0) |
Table to find the coefficient of the general term ((k+1)th term)
| Coefficient of x, A | 2 |
| Constant term B | 3 |
| Exponent n | 5 |
| Term number (from the start) | 3 |
| k (term number minus 1) | =B4-1 |
| Power of x (n−k) | =B3-B5 |
| Binomial coefficient nCk | =COMBIN(B3,B5) |
| Coefficient of the term | =B7*B1^B6*B2^B5 |
Table to find a binomial coefficient (number of combinations)
| n | 5 |
| k | 2 |
| Binomial coefficient nCk | =COMBIN(B1,B2) |
| Check with factorials | =FACT(B1)/(FACT(B2)*FACT(B1-B2)) |
Table to build Pascal's triangle
| Row n = 0 | 1 | |||||
| Row n = 1 | 1 | 1 | ||||
| Row n = 2 | 1 | =B2+C2 | 1 | |||
| Row n = 3 | 1 | =B3+C3 | =C3+D3 | 1 | ||
| Row n = 4 | 1 | =B4+C4 | =C4+D4 | =D4+E4 | 1 | |
| Row n = 5 | 1 | =B5+C5 | =C5+D5 | =D5+E5 | =E5+F5 | 1 |
The same formulas as in Excel work as is (Google Sheets also has COMBIN and FACT with the same names). Copy the whole table, paste it into cell A1, and replace A, B and n with your own numbers.
How to calculate it in Python
from fractions import Fraction
from math import comb
# Expand (Ax + B) to the power n (a fraction such as 3/4 can be written Fraction(3, 4))
coef_x = Fraction(2) # coefficient of x, A
const = Fraction(3) # constant term B
power = 5 # exponent n
# k goes from 0 to n. The power of x is n - k, and the coefficient is nCk × A^(n-k) × B^k
for k in range(power + 1):
exponent = power - k
coefficient = comb(power, k) * coef_x ** exponent * const ** k
print(f"Coefficient of x^{exponent}: {coefficient}")
# Find only the coefficient of x^3 (k = n - 3)
target = 3
k = power - target
coefficient = comb(power, k) * coef_x ** target * const ** k
print(f"Binomial coefficient: {comb(power, k)}")
print(f"Coefficient of x^{target}: {coefficient}")
In the standard library, math.comb gives the binomial coefficient nCk, and fractions.Fraction keeps exact fractions (math.comb needs Python 3.8 or later). This example expands (2x+3)⁵. When you run it, the coefficients appear in the order 32, 240, 720, 1080, 810, 243, and finally the coefficient of x cubed, 720. Change A, B and n and run it.
How to write it in LaTeX and other math languages (copy and paste)
The binomial theorem
(a+b)ⁿ = ₙC₀aⁿ + ₙC₁aⁿ⁻¹b + ⋯ + ₙCₙbⁿ
(a+b)^{n} = \sum_{k=0}^{n} {}_{n}\mathrm{C}_{k}\, a^{n-k} b^{k}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<msup>
<mrow><mo>(</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>)</mo></mrow>
<mi>n</mi>
</msup>
<mo>=</mo>
<munderover>
<mo>∑</mo>
<mrow><mi>k</mi><mo>=</mo><mn>0</mn></mrow>
<mi>n</mi>
</munderover>
<mmultiscripts><mi mathvariant="normal">C</mi><mi>k</mi><none/><mprescripts/><mi>n</mi><none/></mmultiscripts>
<msup><mi>a</mi><mrow><mi>n</mi><mo>−</mo><mi>k</mi></mrow></msup>
<msup><mi>b</mi><mi>k</mi></msup>
</mrow>
</math>
(a+b)^n = sum_(k=0)^n C(n,k) a^(n-k) b^k
Sum[Binomial[n, k] a^(n - k) b^k, {k, 0, n}]
sum(binomial(n, k)*a^(n-k)*b^k, k = 0 .. n);
symsum(nchoosek(n, k)*a^(n-k)*b^k, k, 0, n)
(a+b)^n = ∑_(k=0)^n C(n,k) a^(n-k) b^k
The general term (the \((k+1)\)th term)
Tₖ₊₁ = ₙCₖ aⁿ⁻ᵏ bᵏ
T_{k+1} = {}_{n}\mathrm{C}_{k}\, a^{n-k} b^{k}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<msub><mi>T</mi><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub>
<mo>=</mo>
<mmultiscripts><mi mathvariant="normal">C</mi><mi>k</mi><none/><mprescripts/><mi>n</mi><none/></mmultiscripts>
<msup><mi>a</mi><mrow><mi>n</mi><mo>−</mo><mi>k</mi></mrow></msup>
<msup><mi>b</mi><mi>k</mi></msup>
</mrow>
</math>
T_(k+1) = C(n,k) a^(n-k) b^k
Binomial[n, k] a^(n - k) b^k
T := binomial(n, k)*a^(n-k)*b^k;
T = nchoosek(n, k)*a^(n-k)*b^k;
T_(k+1) = C(n,k) a^(n-k) b^k
Calculating a binomial coefficient (the number of combinations)
ₙCₖ = n! ÷ (k!(n−k)!)
{}_{n}\mathrm{C}_{k} = \frac{n!}{k!\,(n-k)!}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mmultiscripts><mi mathvariant="normal">C</mi><mi>k</mi><none/><mprescripts/><mi>n</mi><none/></mmultiscripts>
<mo>=</mo>
<mfrac>
<mrow><mi>n</mi><mo>!</mo></mrow>
<mrow><mi>k</mi><mo>!</mo><mo>(</mo><mi>n</mi><mo>−</mo><mi>k</mi><mo>)</mo><mo>!</mo></mrow>
</mfrac>
</mrow>
</math>
C(n,k) = (n!)/(k!(n-k)!)
Binomial[n, k]
binomial(n, k);
nchoosek(n, k)
C(n,k) = n!/(k!(n-k)!)
Pascal's triangle (the sum of two neighbors)
ₙ₋₁Cₖ₋₁ + ₙ₋₁Cₖ = ₙCₖ
{}_{n-1}\mathrm{C}_{k-1} + {}_{n-1}\mathrm{C}_{k} = {}_{n}\mathrm{C}_{k}
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
<mrow>
<mmultiscripts><mi mathvariant="normal">C</mi><mrow><mi>k</mi><mo>−</mo><mn>1</mn></mrow><none/><mprescripts/><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow><none/></mmultiscripts>
<mo>+</mo>
<mmultiscripts><mi mathvariant="normal">C</mi><mi>k</mi><none/><mprescripts/><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow><none/></mmultiscripts>
<mo>=</mo>
<mmultiscripts><mi mathvariant="normal">C</mi><mi>k</mi><none/><mprescripts/><mi>n</mi><none/></mmultiscripts>
</mrow>
</math>
C(n-1,k-1) + C(n-1,k) = C(n,k)
Binomial[n - 1, k - 1] + Binomial[n - 1, k] == Binomial[n, k]
binomial(n-1, k-1) + binomial(n-1, k) = binomial(n, k);
nchoosek(n-1, k-1) + nchoosek(n-1, k)
C(n-1,k-1) + C(n-1,k) = C(n,k)
How to have ChatGPT do the calculation
You are an algebra calculation assistant. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing). Use the binomial theorem to expand (2x+3)^5. Show each of the following: 1. The general term (the (k+1)th term) 2. The binomial coefficients 5Ck for k = 0 to 5 3. The expanded expression (in descending powers of x, with each coefficient as a simplified fraction or a whole number) 4. The coefficient of x^3, and how you found it (how you chose k) In Python, calculate exactly with math.comb and fractions.Fraction from the standard library, and show the formulas you used and the numbers from the execution result.
How to Use
-
1Enter your numbersType the numbers you want to calculate with into the input fields
-
2CalculatePress the "Calculate" button
-
3Check the resultThe result appears on the spot. The same page also explains the idea behind the calculation and the formula
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