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Binomial expansion II

Binomial expansion of compound expressions

Binomial expansion of compound expressions

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Summary

​Binomial expansion of compound expressions

​​In a nutshell

The binomial expansion formula allows you to expand expressions of the form (a+bx)n(a + bx)^n(a+bx)n for any real constants aaa​, bbb​ and nnn​. You can use this to compute the expansions of expressions which can be expressed as the product or sum of terms of the form (a+bx)n(a + bx)^n(a+bx)n.



Sums of binomial expansions

If you know the binomial expansions of (a+bx)n(a + bx)^n(a+bx)n​ and (c+dx)m(c + dx)^m(c+dx)m​, then the expansion of (a+bx)n+(c+dx)m(a + bx)^n + (c+dx)^m(a+bx)n+(c+dx)m​ is the sum of their respective expansions. In other words, the coefficient of xkx^kxk​ in the expansion of (a+bx)n+(c+dx)m(a+bx)^n + (c+dx)^m(a+bx)n+(c+dx)m​ is the sum of the coefficients of xkx^kxk​ in the expansions of (a+bx)n(a+bx)^n(a+bx)n​ and (c+dx)m(c+dx)^m(c+dx)m.


The range of values of xxx for which this expansion will hold is the interval for which both the expansions of (a+bx)n(a+bx)^n(a+bx)n​ and (c+dx)m(c+dx)^m(c+dx)m​ hold, so if ∣ab∣≤∣cd∣\left|\dfrac{a}{b}\right| \le \left|\dfrac{c}{d}\right|​ba​​≤​dc​​​ then this is ∣x∣<∣ab∣|x| < \left|\dfrac{a}{b}\right|∣x∣<​ba​​​.

The same holds when adding more than two terms of the form (a+bx)n(a+bx)^n(a+bx)n​ together, and the range of values for which the expansion of the sum will hold is the range of values for which the expansion holds for all of the terms being summed simultaneously.



Example 1

Find the terms up to the x2x^2x2​ term in the expansion of 3x2x2+x−1\dfrac{3x}{2x^2 + x - 1}2x2+x−13x​, and the range of values of xxx for which the expansion holds.


Factorise the quadratic in the denominator to get that 3x2x2+x−1=3x(2x−1)(x+1)\dfrac{3x}{2x^2 + x - 1} = \dfrac{3x}{(2x-1)(x+1)}2x2+x−13x​=(2x−1)(x+1)3x​.


Use partial fractions to find values of AAA and BBB such that 3x(2x−1)(x+1)=A2x−1+Bx+1\dfrac{3x}{(2x-1)(x+1)} = \dfrac{A}{2x-1} + \dfrac{B}{x+1}(2x−1)(x+1)3x​=2x−1A​+x+1B​.


Multiply by the denominator to obtain

3x=A(x+1)+B(2x−1)3x = A(x+1) + B(2x-1)3x=A(x+1)+B(2x−1)​​


Equate coefficients and solve for AAA and BBB​

3x=Ax+A+2Bx−B3=A+2B0=A−BB=A3=3AA=1B=1\begin{aligned}3x &= Ax + A + 2Bx - B\\3 &= A + 2B\\0 &= A - B\\B &= A\\3 &= 3A\\A &= 1\\B &= 1\\\end{aligned}3x30B3AB​=Ax+A+2Bx−B=A+2B=A−B=A=3A=1=1​​​


Therefore 3x2x2+x−1=12x−1+1x+1\dfrac{3x}{2x^2 + x - 1} = \dfrac{1}{2x-1} + \dfrac{1}{x+1}2x2+x−13x​=2x−11​+x+11​.


The expansion of 12x−1=−(1−2x)−1\dfrac{1}{2x-1} = -(1-2x)^{-1}2x−11​=−(1−2x)−1 is:


−(1−2x)−1=−1[1+(−1)(−2x)+(−1)(−2)2!(−2x)2+… ]=−1−2x−4x2−…\begin{aligned}-(1-2x)^{-1} &=-1\left[1 + (-1)(-2x) + \dfrac{(-1)(-2)}{2!}(-2x)^2 + \dots\right]\\&= -1 - 2x - 4x^2 - \dots\end{aligned}−(1−2x)−1​=−1[1+(−1)(−2x)+2!(−1)(−2)​(−2x)2+…]=−1−2x−4x2−…​​​


And this holds when ∣x∣<12|x| < \dfrac12∣x∣<21​.


The expansion of 1x+1=(1+x)−1\dfrac{1}{x+1} = (1+x)^{-1}x+11​=(1+x)−1 is :


​(1+x)−1=[1+(−1)x+(−1)(−2)2!x2+… ]=1−x+x2+…\begin{aligned}(1 + x)^{-1} &= \left[1 + (-1)x + \dfrac{(-1)(-2)}{2!}x^2 + \dots\right]\\&= 1 -x + x^2 + \dots\end{aligned}(1+x)−1​=[1+(−1)x+2!(−1)(−2)​x2+…]=1−x+x2+…​​​

​

And this holds when ∣x∣<1|x| < 1∣x∣<1.


The sum of these is:


​(1−x+x2+… )+(−1−2x−4x2−… )=−3x−3x2−…(1 - x + x^2 + \dots) +(-1 - 2x - 4x^2 - \dots) = -3x -3x^2 - \dots(1−x+x2+…)+(−1−2x−4x2−…)=−3x−3x2−…​​


And this holds when ∣x∣<∣12∣|x| < \left|\dfrac12\right|∣x∣<​21​​


Therefore, the terms up to the x2x^2x2 term in the expansion of 3x2x2+x−1\dfrac{3x}{2x^2 + x - 1}2x2+x−13x​ are −3x−3x2−…‾\underline{-3x - 3x^2 - \dots}−3x−3x2−…​ and this expansion holds when ∣x∣<12|x| < \dfrac12∣x∣<21​.

​


Products of binomial expansions

To compute the expansion of (a+bx)n(c+dx)m(a+bx)^n(c + dx)^m(a+bx)n(c+dx)m up to the xkx^kxk​ term, compute the expansions of both (a+bx)n(a+bx)^n(a+bx)n​ and (c+dx)m(c+dx)^m(c+dx)m​ up to the xkx^kxk term, and take the product of these.


The range of values of xxx for which this expansion holds is the interval for which both the expansions of (a+bx)n(a + bx)^n(a+bx)n​ and (c+dx)m(c+dx)^m(c+dx)m​​ hold, so if ∣ab∣≤∣cd∣\left|\dfrac{a}{b}\right| \le \left|\dfrac{c}{d}\right|​ba​​≤​dc​​​​ then this is ∣x∣<∣ab∣|x| < \left|\dfrac{a}{b}\right|∣x∣<​ba​​.

The same holds when adding more than two terms of the form (a+bx)n(a+bx)^n(a+bx)n together, and the range of values for which the expansion of the sum will hold is the range of values for which the expansion holds for all of the terms being summed simultaneously.



Example 2

Find the coefficient of the x2x^2x2​ term in the expansion of:

 12−xx2+4x+43\sqrt[3]{\dfrac{\dfrac12 - x}{x^2 + 4x + 4}}3x2+4x+421​−x​​ ​


Rewrite the expression as a product of binomials:


12−xx2+4x+43=(12−x)13(2+x)−23\sqrt[3]{\dfrac{\dfrac12 - x}{x^2 + 4x + 4}} = \left(\dfrac12 - x\right)^{\dfrac13}\left(2 + x\right)^{-\dfrac23}3x2+4x+421​−x​​=(21​−x)31​(2+x)−32​​​


Expand these up to the x2x^2x2 term:


(12−x)13(2+x)−23=(12)13(1−2x)13(2)−23(1+12x)−23=12[1+(13)(−2x)+(13)(−23)2!(−2x)2+… ]×[1+(−23)(12x)+(−23)(−53)2!(12x)2+… ]=12[1+−23x−49x2+… ][1+−13x+536x2+… ]\begin{aligned}\left(\dfrac12 - x\right)^{\dfrac13}\left(2 + x\right)^{-\dfrac23} &= \left(\dfrac12\right)^{\dfrac13}\left(1 - 2x\right)^{\dfrac13}\left(2\right)^{-\dfrac23}\left(1 + \dfrac12x\right)^{-\dfrac23}\\&= \dfrac12\left[1 + \left(\dfrac13\right)(-2x) + \dfrac{\left(\dfrac13\right)\left(-\dfrac23\right)}{2!}(-2x)^2 + \dots\right] \\& \qquad \times \left[1 + \left(-\dfrac23\right)\left(\dfrac12x\right) + \dfrac{\left(-\dfrac23\right)\left(-\dfrac53\right)}{2!}\left(\dfrac12x\right)^2 + \dots\right]\\&=\dfrac12\left[1 + -\dfrac23x - \dfrac49x^2 + \dots\right]\left[1 + -\dfrac13x + \dfrac{5}{36}x^2 + \dots\right]\\\end{aligned}(21​−x)31​(2+x)−32​​=(21​)31​(1−2x)31​(2)−32​(1+21​x)−32​=21​​1+(31​)(−2x)+2!(31​)(−32​)​(−2x)2+…​×​1+(−32​)(21​x)+2!(−32​)(−35​)​(21​x)2+…​=21​[1+−32​x−94​x2+…][1+−31​x+365​x2+…]​​​


The x2x^2x2 term of this is therefore:


12(1×536+(−23)×(−13)−49×1)x2=−124x2\dfrac12\left(1 \times\dfrac{5}{36} +\left(-\dfrac23\right)\times\left(-\dfrac13\right) -\dfrac49 \times 1\right)x^2 = -\dfrac{1}{24}x^221​(1×365​+(−32​)×(−31​)−94​×1)x2=−241​x2​​


Therefore, the coefficient of the x2x^2x2 term in the expansion of 12−xx2+4x+43\sqrt[3]{\dfrac{\dfrac12 - x}{x^2 + 4x + 4}}3x2+4x+421​−x​​ is −124‾\underline{-\dfrac{1}{24}}−241​​.



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FAQs - Frequently Asked Questions

What is the coefficient of x^k in the expansion of (a+bx)^n + (c+dx)^m?

The coefficient of x^k​ in the expansion of (a+bx)^n+(c+dx)^m​ is the sum of the coefficients of x^k in the expansions of (a+bx)^n​ and (c+dx)^m.

How do I work out the expansion of a sum of binomials?

If you know the expansions of (a+bx)^n and (c+dx)^m, then the expansion of (a+bx)^n + (c+dx)^m is the sum of their respective expansions.

How do I work out the expansion of a product of binomials?

To compute the expansion of (a+bx)^n(c+dx)^m up to the x^k term, compute the expansions of both (a+bx)^n​ and (c+dx)^m ​ up to the x^k term, and take the product of these.

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