Binomial Theorem
(a + b)n = ∑ [r = 0 → n] nCr · an−r · br
Tr+1 = nCr · an−r · br
nCr = n! ÷ [ r! × (n − r)! ]
Key Results
- Middle term (n even): Single middle term T(n/2)+1, i.e. r = n/2
- Middle terms (n odd): Two middle terms — T(n+1)/2 and T(n+3)/2
- Independent term (constant term): Set total power of x = 0, then solve for r
- Power of x in (xa + x−b)n: Power = a(n − r) − br
- Greatest / numerically largest term: Find r where |Tr+2 / Tr+1| ≥ 1
- Symmetry: nCr = nC(n−r) — always use it to simplify
Sum Identities
- Sum of all coefficients: Put x = 1 ⇒ (a + b)n = 2n
- Pascal’s identity: nCr + nCr−1 = (n+1)Cr
- Inequality: (1 + x)n > 1 + nx for x > 0, n > 1
- Alternating sum (x = −1): nC0 − nC1 + nC2 − … = 0
⚠ Exam Traps — Avoid These
- Sign error in (a − b)n: Use (−b)r — odd r gives negative terms!
- Off-by-one index: T(r+1) is the (r+1)th term — r starts from 0, not 1!
- Expanding fully when a shortcut exists: Always check if the general term Tr+1 is faster.
- Rational / infinite series: (1 + x)n expansion is valid only for |x| < 1 when n ∉ ℤ+
Term Finder — Tr+1
Enter n, r, a, b above then click the button.
Coefficient of xk
Enter values to find the coefficient.
Greatest / Numerically Largest Term in (1 + x)n
Enter n and the value of x.
nCr Calculator
n =
r =
Result will appear here.
Full Expansion — All Terms Listed
Numeric a, b → values computed automatically. Symbolic a, b (e.g. x, 2y) → algebraic expressions shown. Maximum n = 12.
Pascal’s Triangle
Each cell = nCr for that row and position. The highlighted row gives the coefficients of (a + b)n.
Extract Row n — All nCr Values
Enter n to extract all nCr values.
Smart Problem Solver
Type your binomial question in plain English. The solver detects the problem type automatically and shows step-by-step working.
Formula used: T(r+1) = nCr · an−r · br