1. Originally Posted by Tally
$k + 2k + 1 = k(k + 1)(k + 2)$

I add K+1 to both sides and get that? Is that right?
did you see my last hint? adding K + 1 to both sides is wrong. if m is an even integer, the next even integer is m + 2

so, the last even integer in your list is 2n, thus, you must add 2n + 2 to both sides

2. I'm about to leave now, so i'll post a solution. if you want to keep trying, you don't have to look at it. here goes.

We wish to show that the sum of the first n positive, even integers is n(n + 1)

Proof:

We proceed by induction.

Let $P(n):$ " $2 + 4 + \cdots + 2n = n(n + 1)$" for all $n \ge 1$

Then $P(1)$ is true, since $2 = 1(1 + 1)$

Now assume $P(k)$ is true, for some integer $k \ge 1$. We show that $P(k + 1)$ is true.

Since $P(k)$ is true, we have

$2 + 4 + \cdots + 2k = k(k + 1)$

adding the next even integer to both sides, we obtain:

$2 + 4 + \cdots + 2k + (2k + 1) = k(k + 1) + (2k + 1)$

$\Rightarrow 2 + 4 + \cdots + 2k + 2(k + 1) = k(k + 1) + 2(k + 1)$

$\Rightarrow 2 + 4 + \cdots + 2(k + 1) = (k + 1)(k + 2)$

$\Rightarrow 2 + 4 + \cdots + 2(k + 1) = (k + 1)[(k + 1) + 1]$

which is precisely the statement $P(k + 1)$

Thus, $P(n)$ holds by induction.

QED

3. Sorry never saw your post before I posted and then never scrolled up.

Sorry if I am seeming stupid but it's because I am.

So you are saying because 2 was the last even number in original statement then I double the K+1? So if 3 was the last number I would add 3K +3 to both sides?

So if I add 2k+2 to each side I get:

$K + 2k + 2 = k(k + 1)(2k + 2)$

4. Originally Posted by Tally
Sorry never saw your post before I posted and then never scrolled up.

Sorry if I am seeming stupid but it's because I am.

So you are saying because 2 was the last even number in original statement then I double the K+1? So if 3 was the last number I would add 3K +3 to both sides?

So if I add 2k+2 to each side I get:

$K + 2k + 2 = k(k + 1)(2k + 2)$
no.

remember how integers are. when counting, you alternate between odd and even.

so $1,2,3,4,5,6,7,8,... \equiv odd, even, odd, even, odd, even,...$

so you see, you have to skip a number to get from one odd number to the next, likewise for getting from one even number to the next. so consecutive odd and even numbers are always two integers apart. so if n is odd, the next odd integer is n + 2, and if m is even, the next even integer is m + 2. that's where adding 2n + 2 came from. i had to start with the last one and add 2 to get the next one. so it was NOT that i doubled k + 1

and you're not stupid!

5. Thanks for all your help, even though I have learned some news things today I am still wandering.

I feel I need to see similar examples and see the step by step movements. Thank you for all your help. I am sure I will be back tomorrow.

Off to bed, work at 5am, then home at 2pmto take care of kids until 8pm then study before bed again. What did I get myself into lol

6. so, the last even integer in your list is 2n, thus, you must add 2n + 2 to both sides
I understand your solution more now and see it is more about rejigging the components than transfering them into numbers like first step. Yet in your solution you added 2k+1 to each side not 2k+2.

Also when you work with the right side does k still = 1 like in first part?

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