Use weak induction to prove the following statement is true for every positive integer $n$: $2+6+18+\dots+2\cdot 3^{n-1}=3^n-1$ Use weak induction to prove the following statement is true for every positive integer $n$: $$2+6+18+\dots+2\cdot 3^{n-1}=3^n-1$$
Base Step: Prove it is true for $n$.
Inductive Hypothesis: It will be true for $n+1$
What I need to show: That it will be true for n and n+1
Proof Proper:.....
To get this started, how do I prove that it is true for $n$?  What $n$ do I choose?  The fact that there is a "..." in the equation scares me.  How do I know how to quantify $n$ and $n+1$?
 A: The base case is typically the least value of $n$ for which you are proving an assertion applies. Since we are aiming to prove  $$P(n):\quad 2+6+18 + \cdots + 2\cdot 3^{n-1}=3^n-1 \tag{P(n)}$$ holds for all positive integers $n$, we need to prove that $P(n)$ holds for $n \geq 1$, and so we need first to confirm that $P(n)$ holds for the base case $n = 1$.
Base Case: 
$P(1)\; (n = 1): {\bf 2\cdot 3^{1-1}} = 2\cdot 3^0 = 2 = {\bf 3^1 - 1}\tag{base case: True}$
Inductive Hypothesis (IH) 
Now, we assume that $P(k)$ is true for some arbitrary positive integer $n = k$. That is, we assume that it is true that $P(k):\quad 2 + 6 + 18 + \cdots + 2\cdot 3^{k-1}=3^k-1\tag{IH}$
Inductive Step
Now, we will use our inductive hypothesis (IH) to prove that $P(k) \implies P(k+1)$. So our aim is to show, using the inductive hypothesis, that $2 + 6 + 18 + \cdots + 2\cdot 3^{k} =3^{k + 1}-1$:
$$\begin{align} P(k + 1):\quad \underbrace{2 + 6 + 18 + \cdots + 2\cdot 3^{k - 1}}_{= \,\large 3^k - 1,\;by \;\text{IH}} + 2\cdot 3^{k} & = (3^k - 1) + 2\cdot 3^k  \\ \\ 
& = 3\cdot 3^k - 1 \\ \\ & = 3^{k + 1} - 1\end{align}$$
So we have shown, in this third step, that $P(k) \implies P(k+1)$.
Hence, by induction on $n$, we can conclude that $P(n)$ is true for all integers $n \geq 1$
A: Let $P(n): 2+6+18+\cdots+2(3^{n-1})=3^n-1$
So, $P(1):2= 3^1-1$ which is true
Let $P(m)$ is true i.e., $2+6+18+\cdots+2(3^{m-1})=3^m-1$
$P(m+1):$  $2+6+18+\cdots+2(3^{m-1})+2\cdot 3^m=3^m-1+2\cdot 3^m=(1+2)3^m-1=3^{m+1}-1$
So,  $P(m+1)$ will hole true if $P(m)$ is true
But, we have already shown $P(n)$ is true for $n=1$
A: The "base step" isn't to prove that it is true for "$n$"... it is to prove that it is true for the first positive integer.
So, in this case, you want to start off by proving that it is true for $n=1$; that is, that $2=3^1-1$.  (Not too hard of a base case, right?)
Now, the idea is this:  ASSUMING that for some number $n$ you know it is true, show that it is also true for the next number.
A: If you needn't insist on applying the mathematical induction, there exists another proof for this which is more simple.
Let $$S_n=\sum_{k=1}^{n}2 \cdot 3^{k-1}=2+6+18+\dots+2\cdot 3^{n-1}.$$Then, we obtain $$3S_n=\sum_{k=1}^{n}2 \cdot 3^{k}=6+18+54+\dots+2\cdot 3^{n}.$$ The one below minus the above gives that $$2S_n=2 \cdot 3^{n}-2.$$As a result,$$S_n=3^n-1.$$ 
