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Binomial expansion II

Binomial expansion: (1+x)^n

Binomial expansion: (1+x)^n

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Binomial expansion: (1+x)^n

Binomial expansion: (a+bx)^n

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Summary

Binomial expansion: (1+x)n(1+x)^n(1+x)n​​

​​In a nutshell

The binomial expansion formula can be generalised to allow for the computation of expressions of the form (1+x)n(1+x)^n(1+x)n​ when nnn​ is any real number, and ∣x∣<1|x| < 1∣x∣<1 via an infinite series.



Generalised binomial expansion formula

When nnn is a natural number, the binomial expansion of (a+b)n(a + b)^n(a+b)n​ is given by the formula:


​(a+b)n=(n0)an+(n1)an−1b+⋯+(nr)an−rbr+⋯+(nn)bn(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{r}a^{n-r}b^r + \dots + \binom{n}{n}b^n(a+b)n=(0n​)an+(1n​)an−1b+⋯+(rn​)an−rbr+⋯+(nn​)bn​​



Recall that (nr)=n!r!(n−r)!=n×(n−1)×⋯×(n−r+1)r!\binom{n}{r} = \dfrac{n!}{r!(n-r)!} = \dfrac{n \times (n-1) \times \dots \times (n-r+1)}{r!}(rn​)=r!(n−r)!n!​=r!n×(n−1)×⋯×(n−r+1)​​. When nnn​ is a natural number, and r>nr>nr>n​, the expression (nr)=0\binom{n}{r} = 0(rn​)=0​. In the case that nnn is not a natural number however, these binomial coefficients are never equal to 000​ for any natural number rrr​. Consequently, the expansion can be continued in order to obtain an infinite series:


​(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3⋯+n(n−1)(n−2)…(n−r+1)r!xr+…(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3\dots + \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}x^r + \dots(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3⋯+r!n(n−1)(n−2)…(n−r+1)​xr+…​​



This expansion is valid for any n∈Rn \in \mathbb{R}n∈R​ whenever ∣x∣<1|x| < 1∣x∣<1​.


This infinite series can be used to find a valid expansion for (1+bx)n(1+bx)^n(1+bx)n​ when ∣bx∣<1|bx| <1∣bx∣<1​, i.e. when ∣x∣<1∣b∣|x|<\dfrac{1} {|b|}∣x∣<∣b∣1​​.



Example 1

Find the binomial expansion of (1+x)5(1+x)^5(1+x)5 where ∣x∣<1|x| < 1∣x∣<1. 


By following the formula given above, you have

(1+x)5=(50)15x0+(51)14x1+(52)13x2+(53)12x3+(54)11x4+(55)10x5=1+5x+10x2+10x3+5x4+x5‾\begin{aligned}(1+x)^5&=\binom{5}01^5x^0+\binom{5}{1}1^{4}x^1+\binom{5}{2}1^{3}x^2+\binom{5}{3}1^{2}x^3+\binom{5}{4}1^{1}x^4+\binom{5}{5}1^{0}x^5\\&=\underline{1+5x+10x^2+10x^3+5x^4+x^5}\end{aligned}(1+x)5​=(05​)15x0+(15​)14x1+(25​)13x2+(35​)12x3+(45​)11x4+(55​)10x5=1+5x+10x2+10x3+5x4+x5​​​​


Example 2

Find the expansion up to and including the x3x^3x3 term of the following expression:

​(51+x)4(1+x)\dfrac{\left(^5\sqrt{1 + x}\right)^4}{\left(1 + x\right)}(1+x)(51+x​)4​​​

​​

For what values of xxx is the expansion valid?


The expression can be simplified to read:


(1+x)45(1+x)=(1+x)45−1=(1+x)−15\begin{aligned}\dfrac{\left(1 + x\right)^{\frac45}}{\left(1 + x\right)} &= \left(1 + x\right)^{\frac45-1}\\&= \left(1 + x\right)^{-\frac15}\end{aligned}(1+x)(1+x)54​​​=(1+x)54​−1=(1+x)−51​​​​


The generalised binomial expansion formula allows you to compute the expansion of this:


1−15x+(−15)(−65)2!x2+(−15)(−65)(−115)3!x3+…1-\dfrac15x+\dfrac{\left(-\dfrac{1}{5}\right)\left(-\dfrac{6}{5}\right)}{2!}x^2 + \dfrac{\left(-\dfrac15\right)\left(-\dfrac65\right)\left(-\dfrac{11}{5}\right)}{3!}x^3 + \dots1−51​x+2!(−51​)(−56​)​x2+3!(−51​)(−56​)(−511​)​x3+…​​


=1−15x+325x2−11125x3+…= 1 - \dfrac{1}{5}x + \dfrac{3}{25}x^2 - \dfrac{11}{125}x^3 + \dots=1−51​x+253​x2−12511​x3+…​​


The expansion is valid when ∣x∣<1\left|x\right| < 1∣x∣<1.


Therefore, the expansion for (51+x)4(1+x)\dfrac{\left(^5\sqrt{1 + x}\right)^4}{\left(1 + x\right)}(1+x)(51+x​)4​​ up to the x3x^3x3 term is 1−15x+325x2−11125x3+…‾\underline{1 - \dfrac{1}{5}x + \dfrac{3}{25}x^2 - \dfrac{11}{125}x^3 + \dots}1−51​x+253​x2−12511​x3+…​ and is valid for ∣x∣<1‾\underline{|x| < 1}∣x∣<1​.



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FAQs - Frequently Asked Questions

How do I expand (1+x)^n when n is not a natural number?

The binomial expansion formula can be generalised to allow for the computation of expressions of the form (1+x)^n when n is any real number and |x| < 1 via an infinite series.

When is the infinite series expansion for (1+bx)^n valid?

When |x| < 1/|b|

When is the infinite series expansion for (1+x)^n valid?

When |x| < 1.

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