This exercise focuses on applying the Binomial Theorem to expand binomial expressions of the form (a+b)n where n is a positive integer.
For any real numbers a and b and a positive integer n:
(a+b)n=∑r=0n(rn)an−rbr
Expanded form:
(a+b)n=(0n)an+(1n)an−1b+(2n)an−2b2+⋯+(nn)bn
where the binomial coefficients are:
(rn)=r!(n−r)!n!
Binomial coefficients can also be read directly from Pascal's Triangle:
n=0:n=1:n=2:n=3:n=4:111134121361141
Each entry is the sum of the two entries directly above it. The r-th entry in row n gives (rn).
Example 1: Expand (x+2)4
Using the Binomial Theorem with a=x, b=2, n=4:
(x+2)4=(04)x4+(14)x3(2)+(24)x2(2)2+(34)x(2)3+(44)(2)4
=x4+4⋅2x3+6⋅4x2+4⋅8x+16
=x4+8x3+24x2+32x+16
Example 2: Expand (2a−3b)3
Write as (2a+(−3b))3 with n=3:
=(03)(2a)3+(13)(2a)2(−3b)+(23)(2a)(−3b)2+(33)(−3b)3
=8a3+3(4a2)(−3b)+3(2a)(9b2)+(−27b3)
=8a3−36a2b+54ab2−27b3
Example 3: Expand (x+x1)5
With a=x, b=x1, n=5:
=(05)x5+(15)x4⋅x1+(25)x3⋅x21+(35)x2⋅x31+(45)x⋅x41+(55)x51
=x5+5x3+10x+x10+x35+x51
- The expansion of (a+b)n has n+1 terms.
- The general term ((r+1)-th term) is: Tr+1=(rn)an−rbr
- The sum of all binomial coefficients: ∑r=0n(rn)=2n
- The Binomial Theorem applies when n∈N (positive integers).