Prove: 2 + 4 + 6 + ... + 2n= n^2+n for all integers n >= 1.
Proof:
P(n):2+4+6+...+2n= n^2+n
P(2):2+4+6+...+2n= 2^2+2= 6 which is true.
Suppose that P(k): 2+4+6+...+2k= k^2+k where k>=1.
2+4+6+...+2k+(2k+1)= (2k+1)^2+(2k+1)= 4k^2+2k+2k+1 +(2k + 1)= 4k^2+6k+2= 2(2k^2+3k+2)= This is as far as I have gotten
Can someone please help? I am trying to prepare for a Final Exam
Thanks