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.