Pascal's Triangle, binomial coefficients, expanding (a + b)ⁿ, and finding specific terms — with worked examples.
(a + b)ⁿ = Σ C(n,k) · aⁿ⁻ᵏ · bᵏ (k from 0 to n)
n+1 terms total | powers of a and b always sum to n
Binomial Coefficient:
C(n, k) = n! / [k! · (n−k)!]
Also written as ⁿCₖ, nCr, or (n choose k)
Pattern of terms:
Term 1 (k=0): C(n,0)·aⁿ·b⁰ = aⁿ
Term 2 (k=1): C(n,1)·aⁿ⁻¹·b¹ = n·aⁿ⁻¹·b
Term 3 (k=2): C(n,2)·aⁿ⁻²·b²
... continues until k=n ...
Last term (k=n): C(n,n)·a⁰·bⁿ = bⁿ
Rule: Each number = sum of the two numbers directly above it. Row n starts and ends with 1.
Use row 4 of Pascal's Triangle: 1 4 6 4 1
a = x, b = 2, n = 4
= 1·x⁴·2⁰ + 4·x³·2¹ + 6·x²·2² + 4·x¹·2³ + 1·x⁰·2⁴
= x⁴ + 4·2x³ + 6·4x² + 4·8x + 16
= x⁴ + 8x³ + 24x² + 32x + 16
Write as (2a + (−3b))³. Row 3: 1 3 3 1
= 1·(2a)³·(−3b)⁰ + 3·(2a)²·(−3b)¹ + 3·(2a)¹·(−3b)² + 1·(2a)⁰·(−3b)³
= 8a³ + 3·4a²·(−3b) + 3·2a·9b² + (−27b³)
= 8a³ − 36a²b + 54ab² − 27b³
= 8a³ − 36a²b + 54ab² − 27b³
General term: C(n,r)·aⁿ⁻ʳ·bʳ
5th term means r = 4 (since term number = r+1)
a = x, b = −2, n = 8, r = 4
C(8,4) · x^(8−4) · (−2)^4
= 70 · x⁴ · 16
= 1120x⁴
Need (3x)^k · 4^(7−k) where k = 3 (so x³ comes out)
Term: C(7,3) · (3x)³ · 4⁴
= 35 · 27x³ · 256
= 241,920x³
Sum of all C(n,k) for k=0 to n equals 2ⁿ. Row 4 sum: 1+4+6+4+1 = 16 = 2⁴.
C(n,0) − C(n,1) + C(n,2) − ... = 0 for n ≥ 1. Comes from (1 + (−1))ⁿ = 0.
When n is even, the middle term is at k = n/2: C(n, n/2) · a^(n/2) · b^(n/2).
C(n,k) = C(n, n−k). Pascal's Triangle is symmetric. The 3rd term from the left = 3rd from the right.
The Binomial Theorem states that (a + b)ⁿ = Σ C(n,k) · aⁿ⁻ᵏ · bᵏ for k from 0 to n. Each term uses a binomial coefficient C(n,k) = n! / [k!(n−k)!]. The expansion has n+1 terms. The powers of a decrease from n to 0, while the powers of b increase from 0 to n, and the powers always add to n.
Pascal's Triangle gives the binomial coefficients for each expansion. Row 0: 1. Row 1: 1 1. Row 2: 1 2 1. Row 3: 1 3 3 1. Row 4: 1 4 6 4 1. Each number is the sum of the two above it. For (a+b)ⁿ, use row n as the coefficients. Faster than calculating C(n,k) separately for small n.
The general term formula is: Term r+1 = C(n,r) · aⁿ⁻ʳ · bʳ, where r starts at 0. To find the 4th term of (x+2)⁷: use r=3 (since term number = r+1), so Term 4 = C(7,3) · x^(7−3) · 2³ = 35 · x⁴ · 8 = 280x⁴.
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