Question

use mathematical induction to prove that n^2-n is divisible by 2 whenever n is a positive integer

Answer 1

\[let \ n = 1\\1^2-1= 0 \text{ is divisible by 2}\\\text{assume} \ n^2-n \text{ is divisible by 2 for some n}\\then\\(n+1)^2-(n+1)=n^2+2n+1-n-1=n^2+n = n^2+n+n-n\\=n^2-n+2n\]
now we know n^2-n is divisible by 2 so we make write n^2-n = 2k
so we have
\[n^2-n+2n = 2k+2n = 2(k+n)\]this is of course divisible by 2
we have shown that if it is true for some n, then it is true for n+1, and we showed its true for n = 1.
so by mathematical induction it is true for all n.