Given the following formula for $x_n$ in terms of $x_{n-1}$ and $a_{n-1}$. Find a explicit formula for $x_n$ in terms of $x_0$, and all $a_k$.
$$ x_n = \frac{x_{n-1}2^{a_{n-1}} - 1}{3} $$
For example when n = 1, the formula is as follows.
$$ x_1 = \frac{x_{0}2^{a_{0}} - 1}{3} $$
When n=2, it is the following.
$$ x_2 = \frac{x_0 2^{a_0 + a_1} - 2^{a_1} - 3}{9} $$
So, in summary, I want to find an explicit formula for $x_n$ in the general case for any $n$.
I base this suggestion on the cases $n=0,1,2,3$ and assuming that the $a_n$ are at least real/complex numbers such that you can sum them up. Also empty sums where the upper index is smaller than the lower are assumed to be zero, i.e., if $m>n$ we have $\sum_{i=m}^n f_i =0$ for any sequence $(f_i)_{i\in\mathbb{N}}$. Then (to be proven by you)
$$x_n = \dfrac{x_0 2^{\sum_{i=0}^{n-1}a_i}-\sum_{k=0}^{n-1}3^{n-1-k} 2^{\sum_{i=1}^ka_{n-i}}}{3^n}$$
Via induction you get the proof in a few lines because the calculations are straight forward.
