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I currently have an induction proof that I have a little trouble solving:

let $x \in R $ be fixed. Use induction to prove that any $n \in N$,
$$(1+2x)^n = 2^0 \binom{n}{0} x^0 + 2^1\binom{n}{1}x^1 + \dotsb + 2^n\binom{n}{n}x^n = \sum_{i=0}^\infty 2^i\binom{n}{i}x^i$$

Here is what I have so far, I'm not too sure if this is correct but I can't seem to simplify the last step so my guess is that I did something wrong along the way but I'm not too sure what I did wrong.

$\mathtt{Basecase: } \text{for n=1, } (1+2x)^1 = 2^0\binom{1}{0}x^0 + 2^1\binom{1}{1}x^1 = 1+2x$
$\mathtt{Induction} \text{ }\mathtt{ Hyp: } \text{for some n>1,} (1+2x)^{n+1} = \sum_{i=0}^\infty 2^{n+1}\binom{n+1}{n+1}x^{n+1}$
$\mathtt{Induction} \text{ } \mathtt{Step: }$ $$(1+2x)^{n+1} = 2^0\binom{n}{0}x^0+\dotsb+ 2^n\binom{n}{n}x^n + 2^{n+1}\binom{n+1}{n+1}x^{n+1}$$ $$ (1+2x)^{n+1} = (1+2x)^n + 2^{n+1}(1)x^{n+1}$$

I'm not sure where to from here. I can't seem to find any way for the right side to be equal to the left side. Is there a common factor that I'm not seeing here? I apologize for my mis-alligned equal signs, I'm fairly new to using MathJax.

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Induction hypothesis should be:

$$(1+2x)^n = 2^0 \binom{n}{0} x^0 + 2^1\binom{n}{1}x^1 + \dotsb + 2^n\binom{n}{n}x^n = \sum_{i=0}^\infty 2^i\binom{n}{i}x^i$$

Your induction hypothesis and induction step look exactly the same.

And the proof can be found all over the net and in any algebra book, for example here

Your special case has $a=1$ and $b=2x$