Binomial Theorem

Setup

(x + 2)⁴

a-term

b-term

n (1–12)

Enter symbolic terms like x, 2x, 3y, x². n must be an integer 1–12.

Examples

(x + 2)⁶ — 4th term

a-term

b-term

n (1–12)

k (1-indexed)

The kth term = C(n, k−1) · a^(n−k+1) · b^(k−1). k=1 is the first term.

Examples

Pascal's Triangle — Row 4

n (0–12)

Displays Pascal's Triangle from row 0 through row n. Row n is highlighted in gold.

Examples

Result

Enter values above and press Calculate to see the expansion, term, or Pascal's Triangle.

Sigma Notation

—

Final Expansion

—

Terms (k = 0 to n)

The kth Term

—

Formula Applied

—

Binomial Coefficient

C(n,k) = n! / (k!(n−k)!) = (n·(n−1)·…·(n−k+1)) / k!

The binomial theorem states: (a+b)ⁿ = Σ C(n,k)·aⁿ⁻ᵏ·bᵏ for k from 0 to n.

Each coefficient C(n,k) counts the number of ways to choose k items from n — also written as ⁿCₖ or (n choose k).

The multiplicative formula C(n,k) = n·(n−1)·…·(n−k+1) / k! avoids computing large factorials directly and works cleanly for n ≤ 20.

Row n of Pascal's Triangle gives exactly C(n,0), C(n,1), …, C(n,n) — the coefficients of the expansion of (x+1)ⁿ.

To find the kth term (1-indexed): T_k = C(n, k−1) · a^(n−k+1) · b^(k−1). The index r = k−1 is the power on b.

Special Cases

(x+1)ⁿ → coefficients are Pascal's row n directly

(x+1)ⁿ: Since b=1, every b^k = 1. The expansion is C(n,0)xⁿ + C(n,1)xⁿ⁻¹ + … + C(n,n) — just the binomial coefficients as plain numbers.

(x−1)ⁿ: The signs alternate: +, −, +, −, … because b^k = (−1)^k.

Approximation for small x: (1+x)ⁿ ≈ 1 + nx when |x| is very small. This comes from keeping only the first two terms of the expansion.