Prove, disprove, or give a counterexample:
$$\sum_{i=0}^n \left(\frac 3 2 \right)^i = 2\left(\frac 3 2 \right)^{n+1} -2.$$
I went about this as a proof by induction. I did the base case and got the LHS = RHS. When I went to show $P(k) \implies P(k+1)$ I could not get the LHS to equal the RHS. Is this because it isn't a proof by induction? Or, that it cannot be proved?
Any help would be greatly appreciated.
The inductive step:
$$\begin{align}\sum_{i=0}^{k+1}\left(\frac{3}{2}\right)^{i} &= \left(\frac{3}{2}\right)^{k+1} + \sum_{i=0}^{k}\left(\frac{3}{2}\right)^{i}\\ &=\left(\frac{3}{2}\right)^{k+1} + 2\cdot\left(\frac{3}{2}\right)^{k+1} - 2\\ &= 3\cdot\left(\frac{3}{2}\right)^{k+1} - 2\\ &= 3\cdot\frac{2}{3}\left(\frac{3}{2}\right)^{k+2} - 2\\ &= 2\cdot\left(\frac{3}{2}\right)^{(k+1) + 1} - 2\end{align}$$ As an alternative (and arguably, faster) method, we notice that LHS is actually a a geometric series: Let $S$ be defined as $$S = \left(\frac{3}{2}\right)^0 + \left(\frac{3}{2}\right)^1 + \dots \left(\frac{3}{2}\right)^n$$ Then, $$\frac{3}{2}S = \left(\frac{3}{2}\right)^1 + \left(\frac{3}{2}\right)^2 + \dots \left(\frac{3}{2}\right)^{n+1}$$ Substract $S$ from $\frac{3}{2}S$ to get $$\begin{align}\frac{3}{2}S - S &= \left(\frac{3}{2}\right)^{n+1} - \left(\frac{3}{2}\right)^0\\ \frac{1}{2}S &= \left(\frac{3}{2}\right)^{n+1} - 1\\ S &= 2\left(\frac{3}{2}\right)^{n+1} - 2\end{align}$$