I haven't done induction in over a year, but the format of your proof just seems wrong to me. I've been messing with it a bit, and I think this would be more accurate. I substituted 2(k-1)+3 for 2k+1, there wasn't any reason not to and it made it easier to figure out:
First, we define S(k) as a series of additions:
 = 2_0+2_1+2_2+...2_k_-_1+2_k+1)
Then we define S(k+1) as an advancement of that series:
 = 2_0+2_1+2_2+...2_k_-_1+2_k+2_k_+_1+1)
So S(k) can be represented by this equation, with 2 being added k times equal to 2k:
 = 2k+1)
We assume this equation is true for k. Then we see that:
 = S(k)+2)
+1)
Which, when we substitute k+1 for k in S(k), is exactly what we should get. Or something like that anyway; I could never get my head fully around induction since it always seems to be going around in a circle.
So apparently we can only use TeX 3 times in each post? That's kinda unfortunate.
First, we define S(k) as a series of additions:
Then we define S(k+1) as an advancement of that series:
So S(k) can be represented by this equation, with 2 being added k times equal to 2k:
We assume this equation is true for k. Then we see that:
Which, when we substitute k+1 for k in S(k), is exactly what we should get. Or something like that anyway; I could never get my head fully around induction since it always seems to be going around in a circle.
So apparently we can only use TeX 3 times in each post? That's kinda unfortunate.