Introduction

The Binomial Theorem can be generalized to expand expressions where the index 'n' is not a positive integer. It can be any rational number (negative integer or a fraction).

Statement of the Theorem

If n is any rational number, then the expansion of (1+x)n(1+x)^n is an infinite series given by:

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+...(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...

Condition of Validity

Since the expansion is an infinite series, it is only valid (convergent) if the modulus of x is less than 1.

∣x∣<1|x| < 1

Important Standard Expansions

These expansions are frequently used and should be memorized (all are valid for ∣x∣<1|x|<1):

  • (1+x)−1=1−x+x2−x3+...=∑r=0∞(−1)rxr(1+x)^{-1} = 1 - x + x^2 - x^3 + ... = \sum_{r=0}^{\infty} (-1)^r x^r
  • (1−x)−1=1+x+x2+x3+...=∑r=0∞xr(1-x)^{-1} = 1 + x + x^2 + x^3 + ... = \sum_{r=0}^{\infty} x^r
  • (1+x)−2=1−2x+3x2−4x3+...=∑r=0∞(−1)r(r+1)xr(1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + ... = \sum_{r=0}^{\infty} (-1)^r (r+1) x^r
  • (1−x)−2=1+2x+3x2+4x3+...=∑r=0∞(r+1)xr(1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + ... = \sum_{r=0}^{\infty} (r+1) x^r

General Term

The general term, or the (r+1)th(r+1)^{th} term, in the expansion of (1+x)n(1+x)^n is:

Tr+1=n(n−1)(n−2)...(n−r+1)r!xrT_{r+1} = \frac{n(n-1)(n-2)...(n-r+1)}{r!}x^r

Example 1: Basic Expansion

Question: Find the first four terms in the expansion of (1−2x)1/2(1-2x)^{1/2}.

Solution: This is of the form (1+y)n(1+y)^n where y=−2xy=-2x and n=1/2n=1/2. The expansion is valid for ∣−2x∣<1|-2x|<1, i.e., ∣x∣<1/2|x|<1/2. Using the formula 1+ny+n(n−1)2!y2+n(n−1)(n−2)3!y3+...1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + ...:

  • 1st term: 1
  • 2nd term: ny=(1/2)(−2x)=−xny = (1/2)(-2x) = -x
  • 3rd term: (1/2)(1/2−1)2(−2x)2=(1/2)(−1/2)2(4x2)=−12x2\frac{(1/2)(1/2-1)}{2}(-2x)^2 = \frac{(1/2)(-1/2)}{2}(4x^2) = -\frac{1}{2}x^2
  • 4th term: (1/2)(−1/2)(−3/2)6(−2x)3=3/86(−8x3)=−12x3\frac{(1/2)(-1/2)(-3/2)}{6}(-2x)^3 = \frac{3/8}{6}(-8x^3) = -\frac{1}{2}x^3 The expansion is 1−x−12x2−12x3−...1 - x - \frac{1}{2}x^2 - \frac{1}{2}x^3 - ...

Example 2: Finding a Specific Coefficient

Question: Find the coefficient of x5x^5 in the expansion of (1+3x)−4(1+3x)^{-4}.

Solution: This is of the form (1+y)n(1+y)^n where y=3xy=3x and n=−4n=-4. We use the general term formula Tr+1=n(n−1)...(n−r+1)r!yrT_{r+1} = \frac{n(n-1)...(n-r+1)}{r!}y^r. We need the term with y5y^5, so r=5r=5.

T6=(−4)(−5)(−6)(−7)(−8)5!(3x)5T_6 = \frac{(-4)(-5)(-6)(-7)(-8)}{5!}(3x)^5

=(−4)(−5)(−6)(−7)(−8)120(243x5)=6720120(243x5)=56(243x5)=13608x5= \frac{(-4)(-5)(-6)(-7)(-8)}{120}(243x^5) = \frac{6720}{120}(243x^5) = 56(243x^5) = 13608x^5

The coefficient is 13608.

Example 3: Approximation

Question: Find the approximate value of 283\sqrt[3]{28} to four decimal places.

Solution: We write 28 as a number close to a perfect cube: 28=27+1=27(1+1/27)28 = 27+1 = 27(1 + 1/27).

283=(27(1+1/27))1/3=271/3(1+1/27)1/3=3(1+1/27)1/3\sqrt[3]{28} = (27(1+1/27))^{1/3} = 27^{1/3}(1+1/27)^{1/3} = 3(1+1/27)^{1/3}

Now, we expand (1+x)n(1+x)^n with x=1/27x=1/27 and n=1/3n=1/3, which is valid since ∣x∣<1|x|<1. We only need the first few terms.

=3[1+nx+n(n−1)2!x2+...]= 3 \left[ 1 + nx + \frac{n(n-1)}{2!}x^2 + ... \right]

=3[1+13(127)+(1/3)(−2/3)2(127)2+...]= 3 \left[ 1 + \frac{1}{3}(\frac{1}{27}) + \frac{(1/3)(-2/3)}{2}(\frac{1}{27})^2 + ... \right]

=3[1+181−19(1729)+...]≈3[1+0.012345−0.000152]≈3(1.012193)≈3.036579= 3 \left[ 1 + \frac{1}{81} - \frac{1}{9}(\frac{1}{729}) + ... \right] \approx 3 [1 + 0.012345 - 0.000152] \approx 3(1.012193) \approx 3.036579

To four decimal places, the value is 3.0366.