1. ## Math induction explanation

I have a question and have set up a formula, but I am not sure how to do the induction steps. Each time I try I do it wrong. Can someone help? Here is the question:
Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).
Here is my equation:

2+4+6+…+n=n(n+1)

And what I have come up with:

The first step is 2 = 1(1+1); 2=2 which is true.

Now we assume that f(n)=f(n+1) is true

2+4+6+…+n+(n+1)=n(n+1)+(n+1)

Ok, so when I solve it it doesn't work. What did I do wrong???

2. Originally Posted by rbotcaldwell
I have a question and have set up a formula, but I am not sure how to do the induction steps. Each time I try I do it wrong. Can someone help? Here is the question:
Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).
Here is my equation:

2+4+6+…+n=n(n+1) Be careful!! The n'th even number is 2n, not n.

And what I have come up with:

The first step is 2 = 1(1+1); 2=2 which is true.

Now we assume that f(n)=f(n+1) is true

2+4+6+…+n+(n+1)=n(n+1)+(n+1) Should be 2+4+6+…+2n+2(n+1)=n(n+1)+2(n+1).

Ok, so when I solve it it doesn't work. What did I do wrong???
Try taking it from there.

3. ## Huhh?

It still doesnt work. Simplify your equation and it becomes (n+1)(n+2). When you plug into the original equation s(2)=2+4=(2+1)(2+2); 6 does not equal 12. I am so not understanding this and I am sure I am making it harder than needs be!

4. Originally Posted by rbotcaldwell
...Simplify your equation and it becomes (n+1)(n+2)....
exactly! which proves the claim

that's what you get when you replace n with (n + 1) in n(n + 1)

you have shown f(n) => f(n + 1)

also, assuming f(n) = f(n + 1) is not a part of mathematical induction

5. Originally Posted by rbotcaldwell
Now we assume that f(n)=f(n+1) is true

2+4+6+…+n+(n+1)=n(n+1)+(n+1)

Ok, so when I solve it it doesn't work. What did I do wrong???
What has helped me is to visualize the goal. Just write it out, and work towards it. You always want to be able to manipulate your formula to get f(n+1) in the end, if that's the route you're going.

So you say: f(n) implies f(n+1), so in this case your f(n+1) will look like this:

(n+1)(n+2)

which is what you got. So what you are doing by plugging in 2 to f(n+1) is actually getting the answer for 2+1=3, which is 6=1+2+3, which is true.

Induction kind of seems like a hat trick at first, but try proving something with induction that doesn't work for anything but the base case. You'll find it quite impossible.