The binomial theorem expands powers like (2x + 3)⁴ without multiplying brackets repeatedly. Enter a, b and n to see the full expansion, any particular term and the middle term.
Enter a, b and n.
Choose which term to find.
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General term: T(r+1) = C(n, r) (ax)^(n−r) b^r
The binomial theorem expands a power of a two-term expression, such as (3x − 2)⁵, into a sum of terms without multiplying the brackets out five times. This calculator handles any expression of the form (ax + b)ⁿ for whole-number powers up to 20. It lists the full expansion with numerical coefficients, finds any particular term T(r+1) you ask for, and gives the middle term or terms.
The theorem is a Class 11 chapter and a steady source of board and JEE questions: find the fifth term, the coefficient of x³, the middle term, or the term independent of x. It also underpins probability, since binomial probabilities are exactly the terms of (q + p)ⁿ, and approximations such as (1 + x)ⁿ ≈ 1 + nx for small x.
1. Identify a, b and n in (ax + b)ⁿ. Keep the sign with b, so (3x − 2) has b = −2.
2. Write the general term: T(r+1) = C(n, r) × (ax)ⁿ⁻ʳ × bʳ, for r = 0, 1, …, n.
3. Compute each binomial coefficient C(n, r) = n! ÷ (r!(n − r)!), or read it from row n of Pascal's triangle.
4. Multiply by aⁿ⁻ʳ and bʳ to get the numerical coefficient, and attach xⁿ⁻ʳ.
5. There are n + 1 terms. If n is even, the middle term is T(n/2 + 1). If n is odd, there are two: T((n+1)/2) and T((n+3)/2).
6. Check: putting x = 1 in the expansion should give (a + b)ⁿ, the sum of all the coefficients.
Write (ax + b)ⁿ as n brackets multiplied together. Each term of the expansion comes from picking either ax or b from every bracket. A term with b picked r times and ax picked n − r times equals (ax)ⁿ⁻ʳ bʳ. The number of ways to choose which r brackets supply b is C(n, r). Adding all these identical products gives the coefficient C(n, r) aⁿ⁻ʳ bʳ. The same counting explains Pascal's triangle: each entry is the sum of the two above it, because C(n, r) = C(n − 1, r − 1) + C(n − 1, r).
When b is negative, odd powers of b are negative, so the terms alternate in sign. The binomial coefficients themselves are symmetric, C(n, r) = C(n, n − r), and they peak in the middle, which is why the middle term often has the largest coefficient when a and b are equal. Once a and b differ, the largest numerical term can shift away from the middle. Note that term numbering starts at T₁ with r = 0, so the rth term uses r − 1, a frequent source of off-by-one errors.
Because the power of x in T(r+1) is n − r, you can find the coefficient of any power by solving n − r = k for r. For expressions like (x + 1/x)ⁿ the power becomes n − 2r, and setting it to zero gives the term independent of x. This calculator handles only (ax + b)ⁿ with a constant b, so its powers run neatly from n down to 0. For negative or fractional powers, the expansion becomes an infinite series valid only when the second term is small.
Sneha is revising the binomial theorem for her Class 11 exam and wants the full expansion of (3x − 2)⁵, its fourth term, which corresponds to r = 3, and its middle terms.
General term: T(r+1) = C(n, r) × (ax)^(n − r) × b^r
Term T4: C(5, 3) × 3^2 × (-2)^3 = 10 × 9 × -8 = -720 T4 = −720x^2
Full expansion: 243x^5 − 810x^4 + 1,080x^3 − 720x^2 + 240x − 32
Middle term(s): T3 = 1,080x^3, T4 = −720x^2
Answer: Expansion 243x^5 − 810x^4 + 1,080x^3 − 720x^2 + 240x − 32; Term T4 −720x^2; Middle term(s) T3 = 1,080x^3, T4 = −720x^2
Dropping the negative sign of b, which ruins every odd-powered term.
Confusing the rth term with T(r+1): the fourth term uses r = 3.
Raising x to the power but forgetting to raise its coefficient a as well.
Assuming the middle term always has the largest coefficient when a and b are unequal.
Using the finite expansion for negative or fractional powers, which need an infinite series.
Class 11 and JEE questions on general terms, middle terms and specific coefficients.
Computing binomial probabilities, which are terms of (q + p)ⁿ.
Quick approximations such as (1.02)¹⁰ using the first few terms.
Polynomial algebra in engineering and computer science.
Understanding compound growth, since (1 + r)ⁿ expands into simple and compound parts.
How many terms are there?
n + 1 terms.
How do I find the term independent of x?
Pick r so that the power of x, n − r here, is zero.