$$ 3^{n+1}-2^{n+1} = \left [ 38\cdot 2^{n-3}+ \sum_{i=3}^{n}\left ( 2^{n-i}\cdot 3^{i} \right ) \right ],n\geq 3 $$
I had a college entrance exam few days ago, and I checked my answer with others.
They wrote the left one, but I wrote the right one.
I tried to put numbers into it. ( n = 3,4,5,... ), and amazingly the equality holds!
How to prove that both of them are the same?
It's a general fact that: $$x^{n+1} - y^{n+1} = (x - y) (x^n + y x^{n-1} + \dots + y^{n-1} x + y^n) = (x-y) \sum_{i=0}^n y^i x^{n-i}$$
You can prove this by induction, or by expanding and pairing elements.
For $x = 3, y = 2$, you get: $$\begin{align} 3^{n+1} - 2^{n+1} & = (3-2) \sum_{i=0}^n 3^i 2^{n-i} \\ & = 3^0 2^n + 3^1 2^{n-1} + 3^2 2^{n-2} + \sum_{i=3}^n 3^i 2^{n-i} \\ & = 2^3 \cdot 2^{n-3} + 3 \cdot 2^2 \cdot 2^{n-3} + 3^2 \cdot 2 \cdot 2^{n-3} + \sum_{i=3}^n 3^i 2^{n-i} \\ &= 38 \cdot 2^{n-3} + \sum_{i=3}^n 3^i 2^{n-i} \end{align}$$