Induction proof of $1+3+\cdots+3^n=\frac{3^{n+1}-1}{2}$

Question

How would I prove the following by induction?$$1+3+3^2+3^3+\cdots+3^n=\frac{3^{n+1}-1}{2}$$ for all $n\geq 0.$

I kept trying to create a base case but I am not sure how many I need. I also seem to be carrying out the steps incorrectly.

If anyone knows how I can proceed with the following problem, I would greatly appreciate your assistance!

Answer 1

We start the induction with n=0.

It is $3^0=1=\frac{3^{0+1}-1}{2}=\frac{3-1}{2}=1\checkmark$.

Inductive claim:

For arbitrary $n\in\mathbb{N}$ holds $\sum_{k=0}^n 3^k=\frac{3^{n+1}-1}{2}$

Inductive step:

$n\mapsto n+1$

$\sum_{k=0}^{n+1} 3^k=\sum_{k=0}^{n} 3^k+3^{n+1}\stackrel{I.c}{=}\frac{3^{n+1}-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2\cdot 3^{n+1}}{2}=\frac{3\cdot 3^{n+1}-1}{2}=\frac{3^{n+2}-1}{2}\checkmark$

And we are done.

Answer 2

you can get you result like this :
$$ T=1+3+3^2+3^3+\cdots+3^n=S+1 \\ S=3+3^2+3^3+\cdots+3^n \\ {S \over 3} = 1+3+3^2+\cdots+3^{n-1} \\ S - {S \over 3}=3^n-1\\ S = (3^n-1)*3/2={{3^{n+1}-3} \over 2} $$ Result then is :
$$ T = {{3^{n+1}-3} \over 2}+1={{3^{n+1}-1} \over 2} $$

Answer 3

Induction step:

$1+3^1 +3^2+ 3^3+.....3^n+3^{n+1} =$

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

$\dfrac{3^{n+1} +2×3^{n+1}-1}{2} =$

$\dfrac{(1+2)3^{n+1}-1}{2} =$

$\dfrac{3^{n+2}-1}{2}.$