Check out some similar questions!

spanker1 · May 14, 2008, 07:10 PM

prove that if 'n' is an integer then prove that n^2 -n+2 is even

I tried mathematical induction to prove it
I used s of k = k^2-k+2
s of k+1=(K+1)^2-(k+1)+2=k^2+2k+1-k-1+2
=k^2k+k+2
=k(k+1)+2

I don't know how to proceed after this.

Rerepete · May 14, 2008, 07:14 PM

If you look at k, and k+1, they are consecutive numbers, one of which has to be even. Any number multiplied by an even number is even as well.

galactus · May 15, 2008, 05:13 AM

I won't bother showing it's true for 1

If $k^{2}-k+2$ is divisible by 2, then

$(k+1)^{2}-(k+1)+2=k^{2}+2k+1-k-1+2=(k^{2}-k+2)+2k.$

Since $k^{2}-k+2$ is divisible by 2 and 2k is divisible by 2, then

$(k+1)^{2}-(k+1)+2$ is divisible by 2.

robkelly2242 · May 16, 2008, 09:13 PM

If k is odd, k^2 must be odd, so k^2-k is even.
If k is even, k^2 is even, and so is k^2-k.