Question 1080223
You have the right idea. You need to connect the (n+k)th case to the kth case somehow. Here's one way to do just that. I'll start where you left off.


Before I start, I'll refer back to this identity
3 + 5 + 7 +...+ (2k+1) = k(k+2) 
which will be used later. 


Onto the problem. Let's simplify things a bit
3 + 5 + 7 +...+ (2(k+1)+1) = (k+1)((k+1)+2) 
3 + 5 + 7 +...+ (2k+3) = (k+1)(k+3)
3 + 5 + 7 +...+ (2k+3) = k^2+4k+3
3 + 5 + 7 +...+ (2k+1) + (2k+3) = k^2+4k+3


Notice how the portion "3 + 5 + 7 +...+ (2k+1)" is buried in the left side of the last equation above. This works because the (k+1)th case is really just the result of taking the kth case and adding on the next term. 


We can replace all of that with the right side of the identity mentioned earlier.


Take note of the color coding to see how the replacement is happening
3 + 5 + 7 +...+ (2k+1) + (2k+3) = k^2+4k+3
<font color=red>3 + 5 + 7 +...+ (2k+1)</font> + (2k+3) = k^2+4k+3
<font color=red>k(k+2) </font> + (2k+3) = k^2+4k+3


At this point, we just simplify the left side
k(k+2) + (2k+3) = k^2+4k+3
k^2+2k + 2k+3 = k^2+4k+3
k^2+4k+3 = k^2+4k+3
and we have another identity because the two sides are the same. 
We can be more rigorous and subtract x^2 from both sides, subtract 4x from both sides, and subtract 3 from both sides. Doing so will have us end up with 0 = 0 a much clearer identity we know to be always true; however this is more work than needed really.


So this means that if n = k holds true, then n=k+1 also holds true. This domino effect leads to proving the inductive step and proves the original claim to be true for all positive integers, or natural numbers.