Question: > 
> $1 + 3 + 3^2 + ... + 3^{n-1} = \dfrac{3^n - 1}2$
> 
> 
> 

I am stuck at

$\dfrac{3^k - 1}2 + 3^k$

and I'm not sure if I am right or not.

Answer: 
For $n=1$ is true because $1=(3^1-1)/2=2/2=1$.
Suppose that is true for $n\leq k$.
$1+...+3^{k-1}+3^k=\_{HI} \frac{3^k-1}{2}+3^k=\frac{3}{2}3^k-\frac{1}{2}=\frac{3^{k+1}}{2}-\frac{1}{2}=\frac{3^{k+1}-1}{2}$. And then is true for all $n$
**First, show that this is true for $n=1$:**

$\sum\limits\_{k=0}^{1-1}3^k=\frac12(3^1-1)$

**Second, assume that this is true for $n$:**

$\sum\limits\_{k=0}^{n-1}3^k=\frac12(3^n-1)$

**Third, prove that this is true for $n+1$:**

$\sum\limits\_{k=0}^{n}3^k=$

$\color\red{\sum\limits\_{k=0}^{n-1}3^k}+3^n=$

$\color\red{\frac12(3^n-1)}+3^n=$

$\frac12(3^n-1+2\cdot3^n)=$

$\frac12(3\cdot3^n-1)=$

$\frac12(3^{n+1}-1)=$

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Please note that the assumption is used only in the part marked red.