Proof by Mathematical induction!
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)
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!)