Is there a closed form to the equation L = $(\frac{x+a}{x})^n -\left (\frac{x-a}{x}\right)^n$ I have the following equation to resolve:
$$L = \left(\dfrac{x+a}{x}\right)^n  - \left(\dfrac{x-a}{x}\right)^n$$
I've found a similar equation on this question, I couldn't understand line 2 to line 3 : $F_Y(\theta+\epsilon) - F_Y(\theta-\epsilon) = 1-\left(\dfrac{\theta-\epsilon}{\theta}\right)^n$? I believe understanding this I'm able to replicate in my equation.
 A: I'm not quite sure that this is what you want, but this is one of the closed forms.
$$(1+a)^n-(1-a)^n=\sum_{k=0}^{n}\binom{n}{k}a^k-\sum_{k=0}^{n}\binom{n}{k}(-a)^k$$
$$=\sum_{k=0}^{n}\binom{n}{k}\{a^k-(-a)^k\}=2\sum_{k=1}^{\lfloor\frac n2\rfloor}\binom{n}{2k-1}a^{2k-1}.$$
A: Rewrite as
$$L_n=(1+a/x)^n-(1-a/x)^n=(1+u)^n-(1-u)^n$$
to make things cleaner to write.
What does this do: you know the expansion of $(1+u)^n$ into a power series in $u$ (the Binomial theorem). Flipping the sign of $u$ just reflects the function left-to-right, and subtracting the two will kill off all the even terms and double the odd terms (this trick is in fact used many times to symmetrize or antisymmetrize a function).
So the expansion is simply double all the odd terms:
$$L_n=2\sum_{k=1,3,\leq n} {n \choose k} u^k$$
You can't do much more than this, as the original form is already nicely reduced for general $n$. You usually want to go in reverse - take a recursive formula / sequence, and find a closed form. A typical sequence of this type is the Fibbonacci sequence with $F_n=F_{n-1}+F_{n-2}$, which reduces to a very similar formula with two power terms.
This lets you believe there is a recursion formula of second order which generates your sequence for all $n$.
Like this:
$$L_n=2L_{n-1}+L_{n-2}(u^2-1)$$
First two terms have to be given (compute by hand, $L_0=0$, $L_1=2u$), the rest can then be computed from the above.
