Does anybody know where I am going wrong?

Question

Use mathematical induction to prove 2 + 4 + 6 + … +2n = n2 + n holds for all positive integers. This is what I have but is not proving true. I am hoping somebody can see where I am going wrong. Please help!!

The base case is n = 1 so 2 = 1^2 + 1. Assume that the statement is true for n = k. 2 + 4 + 6 + … + 2k = k^2 + k. Show that the statement is true for n = k + 1. (2 + 4 + 6 + … + 2k) + 2(k + 1) = (k + 1)^2 + (k + 1). Substituting (k + 1)^2 + (k + 1) for (2 + 4 + 6 + … +2k), we obtain ((k + 1)^2 + (k + 1)) + 2(k + 1) = (k + 1)^2 + (k + 1).

Answer 1

You've proven the base case.

Next you assume that (2 + 4 + 6 + ... + 2k) is k² + k

Now to get to (2 + 4 + 6 + ... + 2k + 2(k+1) you add the 2(k + 1) term:

k² + k + 2(k + 1)

Pull out a 'k' from the first two terms:
k(k + 1) + 2(k + 1)

Now factor 'k + 1'
(k + 2)( k + 1 )

Let n = k + 1, so you can simplify the equation to:
(n + 1)n

Multiplying through you have:
n² + n

Back in terms of k:
(k + 1)² + (k + 1)

Since you proved it for n = 1, and then assumed if for k and proved k + 1, then you have proven it true by mathematical induction.

I think your mistake was to replace the sum with (k + 1)² + (k + 1), but you should have put in k² + k. You put in the destination, not where you are coming from. :)

Answer 2

You say:

"The base case is n = 1 so 2 = 1^2 + 1. Assume that the statement is true for n = k. 2 + 4 + 6 + … + 2k = k^2 + k. Show that the statement is true for n = k + 1. (2 + 4 + 6 + … + 2k) + 2(k + 1) = (k + 1)^2 + (k + 1)."

So far so good. But then you say "Substituting (k + 1)^2 + (k + 1) for (2 + 4 + 6 + … +2k)" which is wrong. You can substitute k²+k which is what your assumption gives you. So you have
2+4+6...+k+2(k+1) =
k²+k+2(k+1) =
k²+2k+1+k+1 =
(k+1)² + (k+1). QED.

Answer 3

Show the statement is true for n=k+1

We know that:
2 + 4 + 6 + … + 2k = k^2 + k
add 2(k+1) to each side.
2 + 4 + 6 + … + 2k + 2(k+1) = k^2 + k + 2(k+1)
now rewrite it as:
2 + 4 + 6 + … + 2k + 2(k+1) = k^2 + 2k + 1 + k + 1
2 + 4 + 6 + … + 2k + 2(k+1) = (k+1)^2 + (k+1)
2 + 4 + 6 + … + 2k + 2(k+1) = n^2 + n
which is what we wanted to prove.

Answer 4

Where you messed up is this step:

"Substituting (k + 1)^2 + (k + 1) for (2 + 4 + 6 + … +2k), "

The sum of all that is NOT (k + 1)^2 + (k + 1), it's k^2 + k, right?

I could go farther, but I think you know what you're doing. That's a compliment. ;)

Answer 5

Your substitution is wrong. You must add the next term and show that you get the formula but with n replaced with n+1


sum to n+1 = sum to n + next term
=n^2+n +(2(n+1)) = n^2 +3n + 2 = (n+1)^2 +(n+1)

thus sum to n+1 is (n+1)^2+(n+1) as you wanted to show.

Answer 6

existence is all about talking up at the same time as it needs to be achieved. and typically human beings will educate you incorrect besides. The smarter, wiser individual will continually listen at the same time as someone speaks, and could communicate up at the same time as it needs to be achieved. I continually tell myself, "its a lot less annoying to dish it out, yet no longer to take it. at the same time as someone is proving me incorrect, i take it VERY demanding. it truly is complicated yet you need to study from each and every thing and comprehend that it makes you a significantly better individual to have lengthy gone by that journey. it truly is existence. from time to time human beings have a good type of starting to be as a lot as do.