Binomial coefficients, Pascal's Triangle, expanding (a + b)^n, finding specific terms, proof by induction, and key identities — with full worked examples.
Binomial Theorem
(a + b)^n = Σ C(n,k) · a^(n−k) · b^k
k runs from 0 to n
Binomial Coefficient
C(n, k) = n! / [k! · (n−k)!]
also written nCk or (n choose k)
General (k+1)-th Term
T_{k+1} = C(n,k) · a^(n−k) · b^k
k starts at 0, so 1st term has k=0
The binomial coefficient C(n, k) — pronounced "n choose k" — counts the number of ways to select k items from a set of n distinct items when order does not matter. It is the same quantity used in combinations, which is why the Binomial Theorem and combinatorics are deeply connected.
C(n, k) = n! / [k! · (n−k)!]
where n! = n × (n−1) × (n−2) × … × 2 × 1 and 0! = 1
Computed values for reference:
Small values of C(n,k)
C(4, 0) = 1 C(4, 1) = 4
C(4, 2) = 6 C(4, 3) = 4
C(5, 2) = 10 C(6, 3) = 20
C(7, 3) = 35 C(8, 4) = 70
Always-true facts
C(n, 0) = 1 for all n ≥ 0
C(n, n) = 1 for all n ≥ 0
C(n, 1) = n for all n ≥ 1
C(n, k) = C(n, n−k) [symmetry]
Pascal's Identity (the additive rule)
C(n, k) = C(n−1, k−1) + C(n−1, k)
This rule explains how each entry in Pascal's Triangle is the sum of the two entries directly above it. The identity holds because of a combinatorial argument: to choose k items from n, either include item n (then choose k−1 from the remaining n−1) or exclude item n (then choose k from the remaining n−1).
Pascal's Triangle is a triangular array where row n contains all the binomial coefficients C(n, 0) through C(n, n). The triangle is built by the rule that each interior entry is the sum of the two entries directly above it. Row 0 starts with a single 1. To expand (a + b)^n, read the coefficients from row n.
Patterns in the triangle
When to use Pascal vs. the formula
(a + b)^n = C(n,0)·a^n + C(n,1)·a^(n−1)·b + C(n,2)·a^(n−2)·b^2 + … + C(n,n)·b^n
Equivalently: (a + b)^n = Σ [k=0 to n] C(n,k) · a^(n−k) · b^k
Structure of the expansion
Key observations:
a = x, b = 3, n = 4. Use row 4 of Pascal's Triangle: 1 4 6 4 1
= C(4,0)·x^4·3^0 + C(4,1)·x^3·3^1 + C(4,2)·x^2·3^2 + C(4,3)·x^1·3^3 + C(4,4)·x^0·3^4
= 1·x^4·1 + 4·x^3·3 + 6·x^2·9 + 4·x·27 + 1·81
= x^4 + 12x^3 + 54x^2 + 108x + 81
Check: substitute x = 1 to get (1+3)^4 = 256. The result gives 1 + 12 + 54 + 108 + 81 = 256. Correct.
Rewrite as (2x + (−3))^5. Set a = 2x, b = −3, n = 5. Row 5 of Pascal's Triangle: 1 5 10 10 5 1
Term 1 (k=0): C(5,0)·(2x)^5·(−3)^0 = 1·32x^5·1 = 32x^5
Term 2 (k=1): C(5,1)·(2x)^4·(−3)^1 = 5·16x^4·(−3) = −240x^4
Term 3 (k=2): C(5,2)·(2x)^3·(−3)^2 = 10·8x^3·9 = 720x^3
Term 4 (k=3): C(5,3)·(2x)^2·(−3)^3 = 10·4x^2·(−27) = −1080x^2
Term 5 (k=4): C(5,4)·(2x)^1·(−3)^4 = 5·2x·81 = 810x
Term 6 (k=5): C(5,5)·(2x)^0·(−3)^5 = 1·1·(−243) = −243
= 32x^5 − 240x^4 + 720x^3 − 1080x^2 + 810x − 243
Note: when b is negative, the signs alternate — odd k terms are negative, even k terms are positive (since (−3)^k alternates sign).
n = 3, first term = a/2, second term = b. Row 3: 1 3 3 1
= 1·(a/2)^3 + 3·(a/2)^2·b + 3·(a/2)·b^2 + 1·b^3
= a^3/8 + 3·(a^2/4)·b + 3·(a/2)·b^2 + b^3
= a^3/8 + (3a^2 b)/4 + (3a b^2)/2 + b^3
The most common exam question type: find the k-th term, or find the term containing x^m, without expanding the entire expression. This is where the general term formula is essential.
General term formula:
T(r+1) = C(n, r) · a^(n−r) · b^r
r = 0, 1, 2, ..., n (r is the index, term number = r + 1)
6th term means r = 5 (since term number = r + 1, so r = 6 − 1 = 5)
a = x, b = 2, n = 9, r = 5
T_6 = C(9, 5) · x^(9−5) · 2^5
= C(9, 5) · x^4 · 32
= 126 · x^4 · 32
= 4032 x^4
Set a = 2x, b = −5, n = 7. We need (2x)^r to contribute x^3, so we need r = 3.
Wait — power of a in the general term is n − r. So power of (2x) is 7 − r.
We want 7 − r = 3, which gives r = 4.
T_5 = C(7, 4) · (2x)^3 · (−5)^4
= 35 · 8x^3 · 625
= 175,000 x^3
a = x^2, b = 1/x = x^(−1), n = 6.
General term: C(6, r) · (x^2)^(6−r) · (x^(−1))^r
= C(6, r) · x^(12−2r) · x^(−r)
= C(6, r) · x^(12−3r)
For the term to be independent of x (constant term), need 12 − 3r = 0, so r = 4.
T_5 = C(6, 4) · (x^2)^2 · (1/x)^4 → C(6,4) = 15
= 15 (the constant term)
Strategy for "find the term containing x^m"
(1 + x)^n = 1 + nx + C(n,2)x^2 + C(n,3)x^3 + … + x^n
The simplest form of the Binomial Theorem — set a = 1, b = x
This special case is extremely useful. Setting x = 0.01, for instance, gives a quick decimal approximation: (1.01)^10 ≈ 1 + 10(0.01) + 45(0.0001) + … ≈ 1.1046. This form also appears in probability when computing the expansion of (p + q)^n where p + q = 1.
Examples of (1 + x)^n expanded:
(1 + x)^2 = 1 + 2x + x^2
(1 + x)^3 = 1 + 3x + 3x^2 + x^3
(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4
(1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5
C(n,0) + C(n,1) + C(n,2) + … + C(n,n) = 2^n
Proof: set a = 1 and b = 1 in (a + b)^n = 2^n
Verification for n = 5: 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^5. The combinatorial meaning: 2^n counts all subsets of an n-element set, and C(n,k) counts k-element subsets, so adding C(n,k) across all k gives all subsets.
C(n,0) − C(n,1) + C(n,2) − C(n,3) + … = 0 (for n ≥ 1)
Proof: set a = 1 and b = −1 in (a + b)^n = 0^n = 0
Verification for n = 4: 1 − 4 + 6 − 4 + 1 = 0. This means the even-indexed coefficients and odd-indexed coefficients of row n sum to the same value (2^(n−1) each).
Combining identities 1 and 2, the even-position coefficients sum to 2^(n−1):
C(n,0) + C(n,2) + C(n,4) + … = 2^(n−1)
C(n,1) + C(n,3) + C(n,5) + … = 2^(n−1)
For n = 5: even positions: 1 + 10 + 5 = 16 = 2^4. Odd positions: 5 + 10 + 1 = 16 = 2^4.
The binomial coefficient C(n, k) is exactly the same as the combination formula nCk from counting theory. This is not a coincidence — you can derive the Binomial Theorem from a combinatorial argument: when you expand (a + b)^n by multiplying n copies of (a + b) together, the term a^(n−k) b^k arises from choosing which k of the n factors contribute a b. There are C(n, k) such choices.
Combination formula
C(n, k) = nCk = n! / [k! · (n−k)!]
Used in: choosing committee of k from n people, choosing cards from a deck, lottery tickets
Binomial coefficient (same formula)
C(n, k) = n! / [k! · (n−k)!]
Used in: coefficient of a^(n−k) b^k in the expansion of (a + b)^n
Permutations vs. combinations vs. binomial coefficients
nPr = n! / (n−r)! (order matters)
nCr = n! / [r! · (n−r)!] (order does not matter)
C(n,k) = nCk (identical — just different notation)
Relationship: nCr = nPr / r!
See Combinations and Permutations for a full treatment of counting methods that underlie the binomial coefficients.
The Binomial Theorem is proved by the principle of mathematical induction. The proof is a standard example in Stewart's Chapter 12 and is frequently tested. Here is the structure; you should be able to reproduce the key steps.
Step 1: Base Case (n = 1)
For n = 1: (a + b)^1 = a + b. The formula gives C(1,0) a^1 b^0 + C(1,1) a^0 b^1 = a + b. Verified.
Step 2: Inductive Hypothesis
Assume the formula holds for n = m: (a + b)^m = sum of C(m,k) a^(m−k) b^k for k from 0 to m.
Step 3: Inductive Step
Write (a + b)^(m+1) = (a + b)^m · (a + b). Multiply the inductive hypothesis by (a + b) and collect terms with equal powers of b. The coefficient of a^(m+1−k) b^k turns out to be C(m, k−1) + C(m, k). By Pascal's Identity, this equals C(m+1, k). The resulting sum matches the Binomial Theorem for n = m+1.
Step 4: Conclusion
By the principle of mathematical induction, the Binomial Theorem holds for all positive integers n.
Why Pascal's Identity is the key
The inductive step works because C(m, k−1) + C(m, k) = C(m+1, k). This is exactly the rule that each Pascal's Triangle entry equals the sum of the two above it. The whole proof reduces to this one identity. If you can state and use this identity, you can reconstruct the proof.
The Binomial Theorem directly generates the binomial distribution in probability. If an experiment has two outcomes (success with probability p, failure with probability q = 1 − p) and is repeated n times independently, then the probability of exactly k successes is the k-th term of the expansion of (p + q)^n.
P(exactly k successes in n trials) = C(n, k) · p^k · q^(n−k)
where q = 1 − p and (p + q)^n = 1^n = 1 [the probabilities sum to 1]
p = 1/2 (heads), q = 1/2 (tails), n = 6
P(exactly 4 heads) = C(6, 4) · (1/2)^4 · (1/2)^2
= 15 · (1/16) · (1/4)
= 15/64
≈ 0.234 (about 23.4% chance)
p = 0.7, q = 0.3, n = 5
P(all 5 made) = C(5,5) · (0.7)^5 · (0.3)^0 = (0.7)^5 ≈ 0.168
P(exactly 3 made) = C(5,3) · (0.7)^3 · (0.3)^2 = 10 · 0.343 · 0.09 ≈ 0.309
P(0 made) + P(1 made) + … + P(5 made) = (0.7 + 0.3)^5 = 1
The fact that all probabilities sum to 1 is guaranteed by the identity (p + q)^n = 1^n = 1 when p + q = 1. This is a direct consequence of the Binomial Theorem — the sum-of-coefficients identity in disguise.
The 5th term of (a+b)^n has r = 4, not r = 5. The formula T_{r+1} means the term number is one more than r. Always subtract 1 from the term number to get r.
In (2x - 3)^5, a = 2x (not just x) and b = -3 (not just 3). You must compute (2x)^k, which contributes a factor of 2^k. Students often forget to raise the coefficient.
When b is negative, the sign of b^r alternates. b^0, b^2, b^4 are positive; b^1, b^3, b^5 are negative. Do not drop the negative sign from b when computing each term.
The exponent n tells you which row to use, but rows are numbered starting at 0. For (a+b)^5, use row 5 (entries 1, 5, 10, 10, 5, 1), not row 6.
The expansion of (a+b)^n has n + 1 terms, not n terms. (a+b)^4 has 5 terms. The last term is the (n+1)-th term with r = n.
Binomial coefficients use C(n,k) (combinations, order irrelevant). Using permutations P(n,k) gives the wrong answer. On a calculator, press nCr, not nPr.
Level 1 — Expand
Level 2 — Specific terms
Level 3 — Identities
Level 4 — Probability
NailTheTest provides step-by-step solutions and private tutoring for every problem type above.
The Binomial Theorem gives a formula for expanding (a + b)^n without multiplying out n times. It states that (a + b)^n equals the sum from k=0 to n of C(n,k) times a^(n-k) times b^k. You use it whenever you need to expand a binomial to a high power efficiently, find a specific term in an expansion, or work with probability models involving repeated independent events.
The binomial coefficient C(n, k), also written as n choose k or nCk, equals n factorial divided by the product of k factorial and (n minus k) factorial. In symbols: C(n,k) = n! / [k! times (n-k)!]. It counts the number of ways to choose k items from a set of n items without regard to order, which is why it also appears in combinations.
Use the general term formula: the (r+1)-th term of (a+b)^n is C(n,r) times a^(n-r) times b^r. To find the 4th term, set r=3 (since term number equals r+1). For example, the 4th term of (2x - 3)^6 with a=2x and b=-3: C(6,3) times (2x)^3 times (-3)^3 equals 20 times 8x^3 times (-27) equals -4320x^3.
Setting a=1 and b=1 in the Binomial Theorem gives (1+1)^n = 2^n = sum of C(n,k) for k=0 to n. Every entry in row n of Pascal's Triangle gets added once, and the total is always 2^n. For example, row 4 has entries 1, 4, 6, 4, 1 which sum to 16 = 2^4. This result has a combinatorial interpretation: 2^n counts all subsets of an n-element set, and C(n,k) counts the subsets of size k.
The alternating sum C(n,0) - C(n,1) + C(n,2) - C(n,3) + ... equals 0 for any n greater than or equal to 1. This follows from the Binomial Theorem with a=1 and b=-1: (1 + (-1))^n = 0^n = 0. For row 4: 1 - 4 + 6 - 4 + 1 = 0. The signs alternate because each b^k term carries the sign of (-1)^k.
Row n of Pascal's Triangle contains exactly the binomial coefficients C(n,0), C(n,1), ..., C(n,n) in order. Each interior entry is the sum of the two entries directly above it, which corresponds to the identity C(n,k) = C(n-1, k-1) + C(n-1, k). This additive rule makes Pascal's Triangle a fast way to find coefficients for small n without computing factorials.
The proof has two steps. Base case: (a+b)^1 = a + b, and the formula gives C(1,0)a + C(1,1)b = a + b. Check. Inductive step: assume the formula holds for n=m, meaning (a+b)^m equals the sum of C(m,k) times a^(m-k) times b^k. Multiply both sides by (a+b) and regroup terms. Using the Pascal identity C(m,k-1) + C(m,k) = C(m+1,k), the resulting sum matches the formula for n=m+1. Therefore the formula holds for all natural numbers n.
The original page covering Pascal's Triangle, C(n,k), and worked expansions
Full treatment of nCr, nPr, factorial, and counting methods
All Stewart precalculus chapters — complete topic index
Arithmetic, geometric, sigma notation, and series convergence intro
Full proof technique — base case, inductive step, strong induction
Sample spaces, events, conditional probability, and binomial distribution
Interactive practice problems, instant step-by-step solutions, and private tutoring for every topic in Stewart Precalculus. Free to start.
Start Practicing FreeNo account required to try