Vollständige Induktion

sum of k = 1 to n 2 * 3 ^ (k-1) = 3 ^ n -1

Lets see.

For all positive integers

Prove by induction that   $\displaystyle\sum_{k=1}^n\;\;2*3^{k-1}=3^{n}-1$

Step 1:

Prove true for n=1

$LHS=2*3^0=2*1=2\\ RHS=3^{1}-1=3-1=2=LHS\\ \text{So the statement is true for n=1}$

Step 2:

Assume the statement is true foe some positive integer y, then frove true for y+1

that is:

Assume    $\displaystyle\sum_{k=1}^y\;\;2*3^{k-1}=3^{y}-1$

Prove that     $\displaystyle\sum_{k=1}^{y+1}\;\;2*3^{k-1}=3^{y+1}-1$

$\begin{align}\\LHS&=​​​​\displaystyle\sum_{k=1}^{y}\;\;2*3^{k-1}\;\;+\;\;2*3^y\\ &=3^y-1\;\;+\;\;2*3^y\\ &=3*3^y-1\\ &=3^{y+1}-1\\ &=RHS \end{align}\\ \text{Hence if the statement is true for a certain positive integer n=y, it will be true for n=y+1 }\\$

Step 3:

The statement is true for n=1

So it must be true for n=2, n=3 etc

The statement must be true for all positive integer values of n            QED