SOLUTION: I need help with this mathematical induction to show that the given statement is true for all natural numbers, 3 + 5 + 7 +...+ (2n+1) = n(n+2) So far I have the following:

Question 1080223: I need help with this mathematical induction to show that the given statement is true for all natural numbers,
3 + 5 + 7 +...+ (2n+1) = n(n+2)
So far I have the following:
Prove Basis n=1
n(n+2)
= 1(1+2)
= 3
Therefore the statement is true for n=1.
When n=k assume
3 + 5 + 7 +...+ (2n+1) = n(n+2)
replace n with k
3 + 5 + 7 +...+ (2k+1) = k(k+2)
Then n = k+1 must be proven, meaning replace k terms with (k+1)
---------------------------------------------------------------------------------
This is where I get lost, I don't know where/how I should replace with (k+1) despite how simple it sounds
I get
3 + 5 + 7 +...+ (2(k+1)+1) = (k+1)((k+1)+2)
But some sites show a funky set up where the right/first part of the equation gets another (k+1) replacement but don't explain how. I'm sure the left hand side (k+1)((k+1)+2) is correct. I know that after that step you must replace the 3 + 5 + 7 ... etc. but I can't get that far since I'm stuck on the (k+1) replacement! Any help/advice is appreciated and I don't need it solved just some guidance! Thank you!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
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
3 + 5 + 7 +...+ (2k+1) + (2k+3) = k^2+4k+3
k(k+2) + (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.
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