Proof by Mathematical induction!

Hi there,

I have this assignment, and I thought I was on the right path, but when I received it back with suggestions to fix it, I'm still not sure where to go from here. Thanks for your help!

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

Let P be the proposition that the sum of the first n even numbers is n(n+1);
P(n) : 2 + 4 + 6+ … + 2n = n(n+1)

With mathematical induction we first need to show that this is true when n=1
Since we are only discussing even numbers, we will start with 2:
2=n(n+1); n=1 so; 2=1(1+1), following simple algebra we find that this is true for n=1:
2=1(2) and 2=2

Inductive Step: Assume that this is true for n=k; so
P(k) : 2 + 4 + 6 + … +2k =k(k+1)
2+4+6+8+10+12+2(7)=7(7+1)
56 = 56

(HERE I WAS TOLD I NEED TO SOLVE FOR ALL N=K, not just 7....)


Now we need to prove that this is also true for (k+1) whenever P(k) is true,
2 + 4 + 6 + … + 2k + 2(k+1) = (k+1)(k+2)
2+4+6+8+10+12+2(7)+2(7+1) = 7^2 + 3(7) + 2
56 + 16 = 49 + 21 +2
72 = 72

(Here I was told I was wrong again, and it was suggested that I add 2(k+1) to both sides, but I don't see how that works out..... help I'm stuck!)

THanks.

Quote:

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Originally Posted by Conorsmom

Hi there,

I have this assignment, and I thought I was on the right path, but when I received it back with suggestions to fix it, I'm still not sure where to go from here. Thanks for your help!

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

Let P be the proposition that the sum of the first n even numbers is n(n+1);
P(n) : 2 + 4 + 6+ … + 2n = n(n+1)

With mathematical induction we first need to show that this is true when n=1
Since we are only discussing even numbers, we will start with 2:
2=n(n+1); n=1 so; 2=1(1+1), following simple algebra we find that this is true for n=1:
2=1(2) and 2=2

Inductive Step: Assume that this is true for n=k; so
P(k) : 2 + 4 + 6 + … +2k =k(k+1)
2+4+6+8+10+12+2(7)=7(7+1)
56 = 56

(HERE I WAS TOLD I NEED TO SOLVE FOR ALL N=K, not just 7....)


Now we need to prove that this is also true for (k+1) whenever P(k) is true,
2 + 4 + 6 + … + 2k + 2(k+1) = (k+1)(k+2)
2+4+6+8+10+12+2(7)+2(7+1) = 7^2 + 3(7) + 2
56 + 16 = 49 + 21 +2
72 = 72

(Here I was told I was wrong again, and it was suggested that I add 2(k+1) to both sides, but I don't see how that works out..... help I'm stuck!)

THanks.

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You have shown that its true for n=1.

Now, assume it holds for n=k. Then show its true for n=k+1.

Since , we need to show that

Note that

The equation becomes .

Expanding out the left side, we see that







Factoring the left side of the equation, we now end up with .

We have completed the proof by induction.

Does this make sense?

--Chris

Alright, so you got to the inductive step okay.
So you would have:
P(k) = 2 + 4 + 6 + … +2k =k(k+1)

Now, you would have to show that it holds for P(k+1).
P(k+1) = 2 + 4 + 6 + … +2k + 2(k+1) = (k+1)*((k+1)+1)

Simplify:
P(k+1) = 2 + 4 + 6 + … +2k + 2(k+1) = (k+1)*(k+2)

Now for the left hand side, we know 2 + 4 + 6 + ... + 2k = k(k+1), so use substitution and you get:
P(k+1) = k(k+1) + 2(k+1) = (k+1)*(k+2)

Now, work with the left hand side:
k^2 + k + 2k +1
k^2 + 3k + 1

Factor this to get:
(k+1)*(k+2), which is exactly what the right side of the equation is. You have now shown that the property holds for P(k+1) and you are done. I hope this helps.