Let:
- $m>0, n>0, c>0$ be integers
Does it follow that:
$$2^m - 3^n = \begin{cases} -3^n(1 - 2^{-c}) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-c}\right), & m=2n-c\\ \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i}\right), & m=2n\\ 3^n(2^c - 1) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+c}\right), & m=2n+c\\ \end{cases}$$
Is there a more standard way to state these relations? Did I make any mistakes?
Here is my reasoning:
- Case 1: $m=2n$
$$2^{2n} - 3^n = \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i}\right)$$
- Base Case: For $n=2:$
$$2^4 - 3^2 = 3 + 2^2$$
- Inductive Hypothesis:
$$\text{Assume that up to }n \ge 2 \text{ that }2^{2n} - 3^n = \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i}\right)$$
- Inductive Case:
- $2^{2n+2} - 3^{n+1} = 4^{n+1} - 3^{n+1} = 3(4^n - 3^n) +2^{2n} = \sum\limits_{i=0}^{n-1}\left(3^{n-i}2^{2i}\right) + 2^{2n} = \sum\limits_{i=0}^{n} \left(3^{n-i}2^{2i}\right)$
- Case 2: $m < 2n$ with $m = 2n - c$
$$2^{2n-c} - 3^n = -3^n(1 - 2^{-c}) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-c}\right)$$
- Base Case: For $c=1:$
- $2^{2n-1} = \dfrac{1}{2}\left[3^n + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i}\right)\right] = 3^{n}2^{-1} + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-1}\right)$
- $2^{2n-1} - 3^n = -3^{n}2^{-1} + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-1}\right)$
- Inductive Hypothesis:
$$\text{Assume that up to }c \ge 1 \text{ that }2^{2n-c} - 3^n = -3^n(1 - 2^{-c}) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-c}\right)$$
- Inductive Case:
- $2^{2n-(c+1)} = \dfrac{1}{2}\left[3^n + -3^n(1 - 2^{-c}) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-c}\right)\right] = 3^n2^{-(c+1)} + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-(c+1)}\right)$
- $2^{2n-(c+1)} - 3^{n} =-3^n(1 - 2^{-(c+1)}) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-(c+1)}\right)$
- Case 3: $m > 2n$ with $m = 2n + c$
$$2^{2n+c} - 3^n = $3^n(2^c - 1) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+c}\right)$$
- Base Case: For $c=1:$
- $2^{2n+1} = 2\left[3^n + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i}\right)\right] = 3^{n}2 + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+1}\right)$
- $2^{2n-1} - 3^n = 3^n + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+1}\right)$
- Inductive Hypothesis:
$$\text{Assume that up to }c \ge 1 \text{ that }2^{2n+c} - 3^n = 3^n(2^c - 1) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+c}\right)$$
- Inductive Case:
- $2^{2n+(c+1)} = 2\left[3^n + 3^n(2^c-1) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+c}\right)\right] = 3^n2^{c+1} + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i+(c+1)}\right)$
- $2^{2n+(c+1)} - 3^{n} =3^n(2^{c+1}-1) + \sum\limits_{i=0}^{n-1} \left(3^{n-1-i}2^{2i-(c+1)}\right)$
What you've done also looks correct to me. As mathlove's question comment states, you can combine your $3$ cases into $1$ equation by using $m = 2n + d$, with $d$ being an integer (i.e., $-c$, $0$ or $c$ in your $3$ cases), to get
$$2^{2n + d} - 3^{n} = 3^{n}(2^{d} - 1) + \sum_{i=0}^{n-1}(3^{n-1-i}2^{2i+d}) \tag{1}\label{eq1A}$$
To prove \eqref{eq1A}, although induction works as you've done, I believe it's simpler & easier to use that for any real $x$ and $y$, and positive integer $n$, we have
$$x^n - y^{n} = (x - y)\left(\sum_{i=0}^{n-1}x^{n-1-i}y^{i}\right) = (x - y)\left(\sum_{i=0}^{n-1}y^{n-1-i}x^{i}\right) \tag{2}\label{eq2A}$$
In your case, $x = 2^2$ and $y = 3$, giving
$$\begin{equation}\begin{aligned} 2^{2n} - 3^{n} & = (2^2 - 3)\left(\sum_{i=0}^{n-1}3^{n-1-i}(2^2)^{i}\right) \\ 2^{d}\left(2^{2n} - 3^{n}\right) & = 2^{d}\left(\sum_{i=0}^{n-1}3^{n-1-i}2^{2i}\right) \\ 2^{d}\left(2^{2n} - 3^{n}\right) & = \sum_{i=0}^{n-1}3^{n-1-i}2^{2i + d} \\ \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Using this in place of the summation in \eqref{eq1A} means its right side becomes
$$\begin{equation}\begin{aligned} 3^{n}(2^{d} - 1) + 2^{d}\left(2^{2n} - 3^{n}\right) & = 3^{n}2^{d} - 3^{n} + 2^{2n+d} - 3^{n}2^{d} \\ & = 2^{2n+d} - 3^{n} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Since this equals the left side of \eqref{eq1A}, it confirms that equation is always true.