For an induction proof involving sigma starting at 0 can our base case be non-zero?

Question

I am trying to prove the claim to be true for any number n, but I am having a little bit of a problem. If the summation starts from i = 0 can we use 1 for our base case? Because I can see how I can prove it if the base case was 1 but not if it is 0.

Answer 1

Assuming you are proving this statement is true for all non-negative integers, I am finding confusion in why you need to. But, yes, you can just create two base cases $0$ and $1$ if you want to. Using the inductive assumption:

$\sum_{i=0}^{m+1}3^i = \sum_{i=0}^{m+1}3^i + 3^{m+1} = \frac{3^{m+1}-1}{2}+3^{m+1} = \frac{3^{m+1}-1}{2}+\frac{2*3^{m+1}}{2} = \frac{3*3^{m+1}-1}{2} = \frac{3^{m+2}-1}{2}$

which is what we want. But if you want to, just because you want to, then yes it is fine as long as you also have the base case $n=0$ as well.

Answer 2

If you use $n=1$ as base case, and then do the induction proof, you have proven the statement for all $n\geq 1$. If you want to show that the statement holds also for $n=0$, then that needs to be proven as well.

The case $n=0$ reads $$ \sum_{i=0}^0 3^i=\frac{3^{0+1}-1}2 $$ which reduces to $$ 3^0=\frac{3-1}2 $$ I'm sure you can see almost immediately that this is indeed true.