<s>[INST] Proof by induction (exponents) 
Use proof by induction and show that the formula holds for all positive integers:$$1+3+3^2+\dots+3^{n-1}=\frac{3^n -1}2$$

The confusing step in my opinion is the first expression: $3^{n-1}$, when I have to show for $k+1$. Any solutions?
 [/INST] *

*Show $3^{1-1}=\dfrac{3^1-1}{2}$

*Assume $\sum\limits_{k=0}^{n-1}3^k=\dfrac{3^n-1}{2}$

*Prove $\sum\limits_{k=0}^{n}3^k=\dfrac{3^{n+1}-1}{2}$:


*

*$\sum\limits_{k=0}^{n}3^k=(\sum\limits_{k=0}^{n-1}3^k)+3^n$

*$(\sum\limits_{k=0}^{n-1}3^k)+3^n=\dfrac{3^n-1}{2}+3^n$

*$\dfrac{3^n-1}{2}+3^n=\dfrac{3^n-1+2\cdot3^n}{2}$

*$\dfrac{3^n-1+2\cdot3^n}{2}=\dfrac{3^n+2\cdot3^n-1}{2}$

*$\dfrac{3^n+2\cdot3^n-1}{2}=\dfrac{3\cdot3^n-1}{2}$

*$\dfrac{3\cdot3^n-1}{2}=\dfrac{3^{n+1}-1}{2}$
Note that the induction-step is applied only at the second bullet.
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