Knowing binomial expansion of $(1+kx)^n$ shortcut for $(a+kx)^n$

Question

If I've calculated the binomial expansion of $(1+kx)^n$ (where $k$ and $n$ are known) up to some term, is there then a shortcut for calculating $(a+kx)^n$?

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Example: Given that the binomial expansion of $(1+6x)^{\frac23}$ up to and including the term in $x^2$ is $1+4x-4x^2$, what is the binomial expansion of $(8+6x)^{\frac23}$ (up to and including the term in $x^2$)?

Answer 1

Thanks to gammatester's answer and the help I received from Henning Makholm in its comment section, I realised this:

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Let $\ f(x)=(1+kx)^n$, then, because,

$$ (a+kx)^n=\left(a\left(1+\frac{kx}a\right)\right)^n=a^n\left(1+k\frac{x}{a}\right)^n $$ you can calculate

$$ (a+kx)^n=a^nf\left(\frac{x}{a}\right) $$

So basically, each term ($x^m$) in the series has to be divided by $a$ to the same power ($a^m$) and then the whole series must be multiplied by $a$, thus your example becomes:

\begin{align*}(8+6x)^\frac23\approx 8^\frac23+\frac{8^\frac23\cdot 4x}8-\frac{8^\frac23\cdot 4x^2}{8^2}\approx 4+2x-\frac{x^2}4\end{align*}

Answer 2

Hint: $$(a+kx)^n=\left(a(1+kx/a)\right)^n = a^n(1+kx/a)^n$$ Of course this assumes that all powers exist, which would be the case for natural $n,$ but e.g. not for $n=2/3, a=-2.$

Answer 3

As an equivalent alternative to gammatester's answer, you can also just multiply an appropriate power of $a$ onto every term, such that the sum of the powers of $x$ and $a$ is $n$. This is because the general term in the binomial expansion of $(1+kx)^n$ is $$ c1^{n-j}(kx)^j $$ for some $c$, and in the expansion of $(a+kx)^n$ you get terms $$ ca^{n-j}(kx)^j $$ but otherwise with the same constants.

Answer 4

Hint: Let's assume you already know $$f(x)=(1+kx)^n$$

Since \begin{align*} (a+kx)^n=a^n\left(1+k\frac{x}{a}\right)^n\tag{1} \end{align*} you can calculate

\begin{align*} (a+kx)^n=a^nf\left(\frac{x}{a}\right)\tag{2} \end{align*}

Comment: Some details due to OPs comment

• In (1) we use \begin{align*} (a+kx)^n&=\left(\frac{a}{a}\left(a+kx\right)\right)^n\\ &=\left(a\cdot\frac{a+kx}{a}\right)^n\\ &=\left(a\left(1+\frac{kx}{a}\right)\right)^n\\ &=a^n\left(1+k\frac{x}{a}\right)^n\tag{3}\\ \end{align*}

• In (2) we use \begin{align*} a^nf\left(\frac{x}{a}\right)=a^n\left(1+k\frac{x}{a}\right)^n=(a+kx)^n \end{align*} according to (3)

Answer 5

It was enough to begin with

$$(1+x)^q=1+qx+\frac{(q)_{2}}2x^2+\frac{(q)_3}{3!}x^3+\cdots$$

Then you get

$$(a+kx)^q=a^q\left(1+\frac{kx}a\right)^q$$

by substituting $\dfrac{kx}a$ for $x$ and multiplying the whole by $a^q$, giving after simplification of the powers of $a$:

$$(a+kx)^q=a^q+qa^{q-1}kx+\frac{(q)_{2}}2a^{q-2}k^2x^2+\frac{(q)_{3}}{3!}a^{q-3}k^3x^3+\cdots$$

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Note: $(q)_k$ is the "falling factorial" $q(q-1)\cdots(q-k+1)$.