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Can someone verify my proof by induction for the following: $2+6+18+...+2 \cdot 3^{n-1} = 3^n - 1, n\geq 1$.

Basis $n = 1$: \begin{align*} 2+6+18+...+2 \cdot 3^{n-1} = 3^n - 1 &= \sum\limits_{i=1}^{n} 2 \cdot 3^{n-1} = 3^n -1 \\ 2+6+18+...+2 \cdot 3^{1-1} &= 3^1 - 1 \\ \sum\limits_{i=1}^{n} 2 \cdot 3^{1-1} &= 3^1 -1 \\ 2 \cdot 1 = 3-1 \\ 2 = 2 \checkmark \\ \end{align*} Inductive Hypothesis: Assume $P(x)$ is true, i.e., \begin{align*} \sum\limits_{i=1}^{n} 2 \cdot 3^{i-1} &= 3^k -1 \\ \end{align*} Inductive Step: Show $P(k+1)$ is true, i.e., \begin{align*} \sum\limits_{i=1}^{k+1} 2 \cdot 3^{i-1} &= 3^{k+1} -1 \\ \sum\limits_{i=1}^{k+1} 2 \cdot 3^{i} &= 3^{k+1} -1 \\ \end{align*}

Mars Plastic
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GooseDee
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    Recommended reading: How to write a clear induction proof. – JMoravitz Feb 23 '20 at 19:04
  • As for what you have written... it is hard to follow what you think you know, what you don't know, and what was just a lazy typo. For instance, you write $\sum\limits_{i=1}^n 2\cdot 3^{\color{red}{n}-1}$ in your basis step. Not only is it incredibly unnecessary to write out so much text for the basis step, and it being so confusing to see you write $2+6+18+\dots+2\cdot 3^{1-1}$ since in this case the terms $6,18,$ and any terms larger don't appear in the summation... there is definitely something wrong with writing the exponent inside of the summation with an $n$ rather than an $i$ – JMoravitz Feb 23 '20 at 19:07
  • Then in your inductive hypothesis, you are mixing up $x,n,i,k$. Surely, only two of these are needed. Then in your inductive step, you didn't really do anything here... in the one line you wrote what you hope to be true but you haven't shown that it is true. In the next line you seem to have multiplied the LHS by $3$ but did not do anything to change the RHS... – JMoravitz Feb 23 '20 at 19:09
  • Note: $2+6+18+\cdots+2\cdot3^{n-1}=(3-1)+(3-1)\cdot3+(3-1)\cdot3^2+\cdots+(3-1)\cdot3^{n-1}$ telescopes to $3\cdot3^{n-1}-1=3^n-1$ – J. W. Tanner Feb 23 '20 at 20:00

2 Answers2

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You have to adjust the notation, and the inductive step is not proved. The correct formulation is:

$$ P(k): \quad \sum_{i=1}^k 2\cdot 3^{i-1}=3^k-1 $$ $$ P(k+1):\quad \sum_{i=1}^{k+1} 2\cdot 3^{i-1}=\sum_{i=1}^k 2\cdot 3^{i-1} + 2\cdot 3^k = $$ $$ = 3^k-1+2\cdot 3^k= ...... $$

can you complete?

Emilio Novati
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What I would write:

Proposition to be proven $P(n)$ is that $2 + 6 + 18 + ... + 2\cdot3^{n-1} = 3^n-1, \forall n \in \mathbb {Z^+}$, which may also be written $\displaystyle \sum_{i=1}^n 2\cdot 3^{i-1} = 3^n-1, \forall n \in \mathbb{Z^+}$.

Basis step: Consider $P(1)$.

For $n=1, \mathrm{LHS\ (left \ hand \ side)} = 2\cdot 3^{1-1} = 2$

While $\mathrm{RHS} = 3^1 - 1 = 2$.

Hence $\mathrm{LHS} = \mathrm{RHS}$ and $P(1)$ is true.

Inductive step:

Inductive hypothesis: assume $P(k)$ holds for some $k \in \mathbb{Z^+}$. This may be written $\displaystyle \sum_{i=1}^k 2\cdot 3^{i-1} = 3^k-1$ for a particular $k \in \mathbb{Z^+}$.

Now consider $P(k+1)$:

$\mathrm{LHS}_{P(k+1)} \\= \displaystyle \sum_{i=1}^{k+1} 2\cdot 3^{i-1}\\ = 2\cdot 3^{k+1-1} + \displaystyle \sum_{i=1}^{k} 2\cdot 3^{i-1} \\= 2\cdot 3^{k} + \displaystyle \sum_{i=1}^{k} 2\cdot 3^{i-1} \\ = 2\cdot 3^{k} + 3^k - 1 \mathrm{\ (applying\ inductive\ hypothesis)}\\ = 3 \cdot 3^{k} - 1 \\= 3^{k+1} - 1 \\=\mathrm{RHS}_{P(k+1)}$

Hence $P(k) \implies P(k+1)$.

By induction, $P(n)$ is true $\forall n \in \mathbb{Z^+}$ (QED)

Deepak
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