​Binomial expansion of compound expressions

​​In a nutshell

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

Sums of binomial expansions

If you know the binomial 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. In other words, 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 $$(c+dx)^m$$.

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

The same holds when adding more than two terms of the form $$(a+bx)^n$$​ 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 $$x^2$$​ term in the expansion of $$\dfrac{3x}{2x^2 + x - 1}$$, and the range of values of $$x$$ for which the expansion holds.

Factorise the quadratic in the denominator to get that $$\dfrac{3x}{2x^2 + x - 1} = \dfrac{3x}{(2x-1)(x+1)}$$.

Use partial fractions to find values of $$A$$ and $$B$$ such that $$\dfrac{3x}{(2x-1)(x+1)} = \dfrac{A}{2x-1} + \dfrac{B}{x+1}$$.

Multiply by the denominator to obtain

$$3x = A(x+1) + B(2x-1)$$​​

Equate coefficients and solve for $$A$$ and $$B$$​

$$\begin{aligned}3x &= Ax + A + 2Bx - B\\3 &= A + 2B\\0 &= A - B\\B &= A\\3 &= 3A\\A &= 1\\B &= 1\\\end{aligned}$$​​

Therefore $$\dfrac{3x}{2x^2 + x - 1} = \dfrac{1}{2x-1} + \dfrac{1}{x+1}$$.

The expansion of $$\dfrac{1}{2x-1} = -(1-2x)^{-1}$$ is:

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

And this holds when $$|x| < \dfrac12$$.

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

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

​

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

The sum of these is:

​$$(1 - x + x^2 + \dots) +(-1 - 2x - 4x^2 - \dots) = -3x -3x^2 - \dots$$​​

And this holds when $$|x| < \left|\dfrac12\right|$$

Therefore, the terms up to the $$x^2$$ term in the expansion of $$\dfrac{3x}{2x^2 + x - 1}$$ are $$\underline{-3x - 3x^2 - \dots}$$ and this expansion holds when $$|x| < \dfrac12$$.

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Products of binomial expansions

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.

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

The same holds when adding more than two terms of the form $$(a+bx)^n$$ 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 $$x^2$$​ term in the expansion of:

 $$\sqrt[3]{\dfrac{\dfrac12 - x}{x^2 + 4x + 4}}$$ ​

Rewrite the expression as a product of binomials:

$$\sqrt[3]{\dfrac{\dfrac12 - x}{x^2 + 4x + 4}} = \left(\dfrac12 - x\right)^{\dfrac13}\left(2 + x\right)^{-\dfrac23}$$​​

Expand these up to the $$x^2$$ term:

$$\begin{aligned}\left(\dfrac12 - x\right)^{\dfrac13}\left(2 + x\right)^{-\dfrac23} &= \left(\dfrac12\right)^{\dfrac13}\left(1 - 2x\right)^{\dfrac13}\left(2\right)^{-\dfrac23}\left(1 + \dfrac12x\right)^{-\dfrac23}\\&= \dfrac12\left[1 + \left(\dfrac13\right)(-2x) + \dfrac{\left(\dfrac13\right)\left(-\dfrac23\right)}{2!}(-2x)^2 + \dots\right] \\& \qquad \times \left[1 + \left(-\dfrac23\right)\left(\dfrac12x\right) + \dfrac{\left(-\dfrac23\right)\left(-\dfrac53\right)}{2!}\left(\dfrac12x\right)^2 + \dots\right]\\&=\dfrac12\left[1 + -\dfrac23x - \dfrac49x^2 + \dots\right]\left[1 + -\dfrac13x + \dfrac{5}{36}x^2 + \dots\right]\\\end{aligned}$$​​

The $$x^2$$ term of this is therefore:

$$\dfrac12\left(1 \times\dfrac{5}{36} +\left(-\dfrac23\right)\times\left(-\dfrac13\right) -\dfrac49 \times 1\right)x^2 = -\dfrac{1}{24}x^2$$​​

Therefore, the coefficient of the $$x^2$$ term in the expansion of $$\sqrt[3]{\dfrac{\dfrac12 - x}{x^2 + 4x + 4}}$$ is $$\underline{-\dfrac{1}{24}}$$.