Binomial Theorem Formula
For any positive integer n:
(a+b)n=∑k=0n(kn)an−kbk
Expanded:
(a+b)n=(0n)an+(1n)an−1b+(2n)an−2b2+⋯+(nn)bn
Binomial Coefficients
(kn)=k!(n−k)!n!=C(n,k)=nCk
Properties
(0n)=(nn)=1
(1n)=(n−1n)=n
(kn)=(n−kn)
(kn)+(k+1n)=(k+1n+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 n gives the coefficients for (a+b)n
Common Expansions
(a+b)2
(a+b)2=a2+2ab+b2
(a−b)2
(a−b)2=a2−2ab+b2
(a+b)3
(a+b)3=a3+3a2b+3ab2+b3
(a−b)3
(a−b)3=a3−3a2b+3ab2−b3
(a+b)4
(a+b)4=a4+4a3b+6a2b2+4ab3+b4
(a+b)5
(a+b)5=a5+5a4b+10a3b2+10a2b3+5ab4+b5
Finding a Specific Term
The (k+1)th term in (a+b)n:
Tk+1=(kn)an−kbk
Example
Find the 4th term of (2x+3)5:
- k=3 (4th term means k+1=4)
- T4=(35)(2x)5−3(3)3
- T4=10⋅4x2⋅27=1080x2
Binomial Series (for any real n)
When ∣x∣<1:
(1+x)n=1+nx+2!n(n−1)x2+3!n(n−1)(n−2)x3+⋯
Special Cases
Sum of Coefficients
(1+1)n=2n=sum of all binomial coefficients in row n
Alternating Sum
(1−1)n=0=(0n)−(1n)+(2n)−(3n)+⋯
Middle Term
For (a+b)n:
- If n is even: One middle term at position (n/2+1)
- If n is odd: Two middle terms at positions (n+1)/2 and (n+3)/2
Applications
- Probability: (kn) gives number of ways to choose k items from n
- Approximations: (1+x)n≈1+nx for small x
- Combinatorics: Counting problems
- Polynomial expansion: Simplifying expressions