# Thread: Proof by Mathematical induction!

1. ## 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.

2. 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.
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 $2+4+6+8+...+2k=k(k+1)$, we need to show that $2+4+6+8+...+2k+2(k+1)=(k+1)(k+2)$

Note that $\underbrace{2+4+6+8+...+2k}_{k(k+1)}+2(k+1)=(k+1)( k+2)$

The equation becomes $k(k+1)+2(k+1)=(k+1)(k+2)$.

Expanding out the left side, we see that

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

$\implies k^2+k+2k+2=(k+1)(k+2)$

$\implies k^2+3k+2=(k+1)(k+2)$

Factoring the left side of the equation, we now end up with $(k+1)(k+2)=(k+1)(k+2)\blacktriangleleft$.

We have completed the proof by induction.

Does this make sense?

--Chris

3. ## Solution

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.