Answer
See proof
Work Step by Step
We have to prove the statement:
$2+6+18+.....+2\cdot 3^{n-1}=3^n-1$
Step 1: Prove that the statement is true for $n=1$:
The left side is:
$2\cdot 3^{1-1}=2$
The right side is:
$3^1-1=2$
As $2=2$, the statement is true for $n=1$.
Step 2: We have to prove:
$2+6+18+.....+2\cdot 3^{k-1}+2\cdot 3^k=3^{k+1}-1$
Because the statement is true for $k$ we have:
$2+6+18+.....+2\cdot 3^{k-1}=3^k-1$
Add $2\cdot 3^k$ on each side:
$2+6+18+.....+2\cdot 3^{k-1}+2\cdot 3^k=3^k-1+2\cdot 3^k$
$2+6+18+.....+2\cdot 3^{k-1}+2\cdot 3^k=3^k(1+2)-1$
$2+6+18+.....+2\cdot 3^{k-1}+2\cdot 3^k=3^{k+1}-1$
We found that the statement is also true for $k+1$.
So, the statement is $\fbox{true}$.
