Introductory mathematical induction problem

[SYoungblood]

Homework Statement

I am just learning the joys of mathematical induction, and this problem is giving me fits.

Homework Equations

I am trying to prove that 2 + 4 + 6 + … + 2n = [2n(n+1)]/2

The Attempt at a Solution

The base case is to prove P(1) is correct. Simple enough -- 2 = [2 x 1 (1+1)]/2. The RHS does equal 2, so we are good to go there.

Next, substitute k for n. So, now we have 2 + 4 + 6 + … + 2k = [2k(k+1)]/2

We have to replace k with (k + 1), which is what we need to prove in this proof. We can also rewrite the LHS to show 2k + 2(k+1) = [2k(k+1)]/2 -- I suspect this is the step where I have gone off the rails.

With a little bit of algebra, I should be able to multiply the equations pout and prove the LHS = RHS, but I am somehow missing something, and I am pretty sure it is in step 3. The example problems I have followed in my text show the algebra is similar problems being pretty straightforward, but I am just not getting how to rewrite the substitution for k with (k + 1) as something mathematically correct.

Thank you for your help in advance.

 

[SammyS]

Homework Statement

I am just learning the joys of mathematical induction, and this problem is giving me fits.

Homework Equations

I am trying to prove that 2 + 4 + 6 + … + 2n = [2n(n+1)]/2

The Attempt at a Solution

The base case is to prove P(1) is correct. Simple enough -- 2 = [2 x 1 (1+1)]/2. The RHS does equal 2, so we are good to go there.

Next, substitute k for n. So, now we have 2 + 4 + 6 + … + 2k = [2k(k+1)]/2

Right. That's what you assume is true for some k where k ≥ 1.

We have to replace k with (k + 1), which is what we need to prove in this proof. We can also rewrite the LHS to show 2k + 2(k+1) = [2k(k+1)]/2 -- I suspect this is the step where I have gone off the rails.

With a little bit of algebra, I should be able to multiply the equations pout and prove the LHS = RHS, but I am somehow missing something, and I am pretty sure it is in step 3. The example problems I have followed in my text show the algebra is similar problems being pretty straightforward, but I am just not getting how to rewrite the substitution for k with (k + 1) as something mathematically correct.

Thank you for your help in advance.

With the above assumption, you now have to prove that your statement is true for k + 1 .

In other words, (with the above assumption) you need to prove that 2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2 .

 

[SYoungblood]

Right. That's what you assume is true for some k where k ≥ 1.With the above assumption, you now have to prove that your statement is true for k + 1 .

In other words, (with the above assumption) you need to prove that 2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2 .

OK, we can assume all of the constants to be true, and remove them form the proof. So, I now have

2k + 2(k+1) = [2(k + 1)(k +1 +1)]/2

So,

2k + 2(k + 1) = (k + 1) (k + 2)
2

Right. That's what you assume is true for some k where k ≥ 1.With the above assumption, you now have to prove that your statement is true for k + 1 .

In other words, (with the above assumption) you need to prove that 2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2 .

Right. That's what you assume is true for some k where k ≥ 1.With the above assumption, you now have to prove that your statement is true for k + 1 .

In other words, (with the above assumption) you need to prove that 2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2 .

So, if I assume the constants to be true and can drop them from the proof, I have 2k + 2(k +1) = [2(k + 1)(k +1 +1)]/2

So,

2k + 2(k +1) = [2(k + 1)(k +1 +1)]/2
2k + 2k + 2 = (k + 1) (k + 2)
4k + 2 = k^2 + 3k + 2

And here is the step that has been derailing me, in some way, shape, or form, since I started working on this thing -- what am I doing wrong here?

 

[SammyS]

OK, we can assume all of the constants to be true, and remove them form the proof. So, I now have

2k + 2(k+1) = [2(k + 1)(k +1 +1)]/2

So,

2k + 2(k + 1) = (k + 1) (k + 2)
2
So, if I assume the constants to be true and can drop them from the proof, I have 2k + 2(k +1) = [2(k + 1)(k +1 +1)]/2

So,

2k + 2(k +1) = [2(k + 1)(k +1 +1)]/2
2k + 2k + 2 = (k + 1) (k + 2)
4k + 2 = k^2 + 3k + 2

And here is the step that has been derailing me, in some way, shape, or form, since I started working on this thing -- what am I doing wrong here?

First of all, what are you calling a "constant" ?

You can't assume that the following is true!

2 + 4 + 6 + … + 2k + 2(k + 1) = [2(k + 1)(k +1 +1)]/2​

You have to show (prove) that it's true assuming that the following is true. You have to show that it's a logical consequence of the following being true.

2 + 4 + 6 + … + 2k = [2k(k+1)]/2​

One plan of attack is to start with 2 + 4 + 6 + … + 2k = [2k(k+1)]/2 and then add 2(k+1) to both sides.