Advanced AlgebraTopic #16 of 32

The Binomial Theorem

Expansion of binomial expressions and Pascal's triangle.

Binomial Theorem Formula

For any positive integer nn:

(a+b)n=k=0n(nk)ankbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

Expanded:

(a+b)n=(n0)an+(n1)an1b+(n2)an2b2++(nn)bn(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

Binomial Coefficients

(nk)=n!k!(nk)!=C(n,k)=nCk\binom{n}{k} = \frac{n!}{k!(n-k)!} = C(n,k) = {}_nC_k

Properties

(n0)=(nn)=1\binom{n}{0} = \binom{n}{n} = 1

(n1)=(nn1)=n\binom{n}{1} = \binom{n}{n-1} = n

(nk)=(nnk)\binom{n}{k} = \binom{n}{n-k}

(nk)+(nk+1)=(n+1k+1)\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}

Pascal's Triangle

Each number is the sum of the two numbers above it:

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
Row 5:      1 5 10 10 5 1
Row 6:     1 6 15 20 15 6 1

Row nn gives the coefficients for (a+b)n(a + b)^n

Common Expansions

(a+b)2(a + b)^2

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

(ab)2(a - b)^2

(ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2

(a+b)3(a + b)^3

(a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

(ab)3(a - b)^3

(ab)3=a33a2b+3ab2b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

(a+b)4(a + b)^4

(a+b)4=a4+4a3b+6a2b2+4ab3+b4(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

(a+b)5(a + b)^5

(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

Finding a Specific Term

The (k+1)(k+1)th term in (a+b)n(a + b)^n:

Tk+1=(nk)ankbkT_{k+1} = \binom{n}{k} a^{n-k} b^k

Example

Find the 4th term of (2x+3)5(2x + 3)^5:

  • k=3k = 3 (4th term means k+1=4k+1 = 4)
  • T4=(53)(2x)53(3)3T_4 = \binom{5}{3}(2x)^{5-3}(3)^3
  • T4=104x227=1080x2T_4 = 10 \cdot 4x^2 \cdot 27 = 1080x^2

Binomial Series (for any real n)

When x<1|x| < 1:

(1+x)n=1+nx+n(n1)2!x2+n(n1)(n2)3!x3+(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots

Special Cases

Sum of Coefficients

(1+1)n=2n=sum of all binomial coefficients in row n(1 + 1)^n = 2^n = \text{sum of all binomial coefficients in row } n

Alternating Sum

(11)n=0=(n0)(n1)+(n2)(n3)+(1 - 1)^n = 0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots

Middle Term

For (a+b)n(a + b)^n:

  • If nn is even: One middle term at position (n/2+1)(n/2 + 1)
  • If nn is odd: Two middle terms at positions (n+1)/2(n+1)/2 and (n+3)/2(n+3)/2

Applications

  1. Probability: (nk)\binom{n}{k} gives number of ways to choose kk items from nn
  2. Approximations: (1+x)n1+nx(1 + x)^n \approx 1 + nx for small xx
  3. Combinatorics: Counting problems
  4. Polynomial expansion: Simplifying expressions