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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.

Community
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Vincent
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  • possible duplicate of Proofs Homework Help – ml0105 Mar 15 '14 at 00:29
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    Well it is. I got an earful about using poor notation, posting technique and two questions in one post. I thought doing it over correctly would be better. – Vincent Mar 15 '14 at 00:30
  • It can be proved by induction, so you must not have seen the algebraic steps necessary to see that the two sides are equal. (It can also be proved by appealing to the known formula for the sum of a geometric series.) – Greg Martin Mar 15 '14 at 01:38

1 Answers1

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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 $$\begin{align}\frac{3}{2}S - S &= \left(\frac{3}{2}\right)^{n+1} - \left(\frac{3}{2}\right)^0\\ \frac{1}{2}S &= \left(\frac{3}{2}\right)^{n+1} - 1\\ S &= 2\left(\frac{3}{2}\right)^{n+1} - 2\end{align}$$

Yiyuan Lee
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