# Induction Questions

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• Mar 3rd 2008, 05:26 AM
Tally
Induction Questions
I have been set the question below, it has been 20+ years since I did all of this so I am a little bit rusty. A place to start would be good, and I will try to work my way to the end. I just need to spark the old cells.

Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).

• Mar 4th 2008, 07:05 AM
Tally
Okay giving it a go, please tell me where I went wrong. (Do you like the confidence?)(Giggle)

Let p(n) be the statement 2+4+6+...+n=n(n+1)

2=1(1+1) Basic step... That can't be right because if I use 2 for the first n, wouldn't I have to use it for all the n's? Or because the questions uses the first even number is it correct?

Help please before I try to move on and really mess it up.
• Mar 4th 2008, 01:50 PM
Jhevon
Quote:

Originally Posted by Tally
Okay giving it a go, please tell me where I went wrong. (Do you like the confidence?)(Giggle)

Let p(n) be the statement 2+4+6+...+n=n(n+1)

2=1(1+1) Basic step... That can't be right because if I use 2 for the first n, wouldn't I have to use it for all the n's? Or because the questions uses the first even number is it correct?

Help please before I try to move on and really mess it up.

your problem has to be more specific. you have to say, "the first n non-negative even integers" or "the first n positive even integers" or so.

but let's say it is as you interpreted, and it is the sum of n even integers starting from 2.

then, let $\displaystyle P(n):$ "$\displaystyle 2 + 4 + 6 + \cdots + 2n = n(n + 1)$" for all integers $\displaystyle n \ge 1$

Now, show that $\displaystyle P(1)$ holds.

Then, assume $\displaystyle P(n)$ holds, and show $\displaystyle P(n)$ implies $\displaystyle P(n + 1)$

then your induction proof will be complete
• Mar 4th 2008, 01:59 PM
Tally
Thank you very much for your assistance. So you are saying that the initial question needed to be more specific or the statement that I made?

So it is 2n because 2 was first number in statement?

Again thanks, I'm sure I will be back. (Wink)
• Mar 4th 2008, 02:03 PM
Jhevon
Quote:

Originally Posted by Tally
Thank you very much for your assistance. So you are saying that the initial question needed to be more specific or the statement that I made?

yes, the initial problem must be more specific. remember, there are an infinite number of even integers, and they go in both directions, negative and positive. we need a direction for our question, or we can't use induction.
Quote:

So it is 2n because 2 was first number in statement?

Again thanks, I'm sure I will be back. (Wink)
no, it is 2 n because the nth even integer is 2n.

example. speaking about the positive, even integers, which is the 2nd one? well, it's 4. what is 4? 4 = 2*2. what's the 3rd? 6, and 6 = 2*3. what's the 4th? it is 8, and 8 = 2*4. in general then, the nth even integer is 2*n, and that is why my sum on the left hand side ends with 2n.

can you complete the problem now?
• Mar 4th 2008, 02:10 PM
Tally
I believe so yes, I will post it later in here to see if it is correct. If you got time tomorrow please look to see if I am right, thanks again. :)

Is the voting over? (Evilgrin)(Rofl)
• Mar 4th 2008, 02:15 PM
Jhevon
Quote:

Originally Posted by Tally
I believe so yes, I will post it later in here to see if it is correct. If you got time tomorrow please look to see if I am right, thanks again. :)

ok. i'll look out for your response

Quote:

Is the voting over? (Evilgrin)(Rofl)
yes, a LLOOONNNGGG time ago, hehe

take care
• Mar 4th 2008, 03:32 PM
Tally
$\displaystyle P(n):2 + 4 + 6 + ... + 2n = n(n + 1)\forall \ge 1$

$\displaystyle \begin{array}{l} 2 \times 1 = 1(1 + 1) \\ 2 = 1(1 \times 1) \\ 2 = 1 \times 2 \\ 2 = 2 \\ \end{array}$
Basic Step P(n) is true.

Now to show that P(N+1) is true.

$\displaystyle 2n + (n + 1) = n(n + 1)(n + 1)$
$\displaystyle \begin{array}{l} 2 + 2 = 1(2)(2) \\ 4 = 2(2) \\ 4 = 4 \\ \end{array}$

P(n+1) is true.

Thank you just starting will add rest soon. Feeding and bathing kids. Not at same time thought even if it would save time.
I'm sure there is an equation for that. (Cool)
• Mar 4th 2008, 03:33 PM
Jhevon
Quote:

Originally Posted by Tally
$P(n):2 + 4 + 6 + ... + 2n = n(n + 1)\forall \ge 1$

you have to change the  tags to  tags
• Mar 4th 2008, 04:34 PM
Tally
That was too easy right?(Thinking)
• Mar 4th 2008, 04:46 PM
Jhevon
Quote:

Originally Posted by Tally
That was too easy right?(Thinking)

what you did was incorrect, i'm afraid.

Here is your objective:

you want to start with the statement $\displaystyle 2 + 4 + \cdots + 2n = n(n + 1)$

and you want to end up with the statement: $\displaystyle 2 + 4 + \cdots + 2(n + 1) = (n + 1)(n + 2)$
• Mar 4th 2008, 05:24 PM
Tally
Thats what I had written down on my paper. I deleted it because I wasn't sure how to get there and thought it looked wrong. Took the easy route, okay back to the scribble board.
(Doh)(Time)
• Mar 4th 2008, 05:40 PM
Tally
Having a harder time than I thought with the this. Is there somewhere that explains the inductive stage in real basic terms because I am not grasping the material that I am reading.(Headbang)
• Mar 4th 2008, 05:53 PM
Jhevon
Quote:

Originally Posted by Tally
Having a harder time than I thought with the this.

here is a hint: if m is an even integer, then the next even integer is (m + 2)

can you go anywhere with that?
Quote:

Is there somewhere that explains the inductive stage in real basic terms because I am not grasping the material that I am reading.(Headbang)
well, the general induction method is what we are doing. to be more specific though is really something to consider by cases. there is not a one-size-fits-all method to show that P(n) implies P(n + 1). how you do that is dependent upon what you are trying to prove

and, of course, just to be explicit, P(n + 1) is the statement you would get if you replaced n with (n + 1) in the P(n) statement
• Mar 4th 2008, 06:27 PM
Tally
$\displaystyle K + K+1 = k(k + 1)(k + 2)$

I add K+1 to both sides and get that? Is that right?
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