Prove by Mathenatical induction that n (n+1) is even

step 1 p(n) = 1 
1 (1+1) = 1 x 2 = 2 
p (1) is true 
step 2 let p(k) = k  (k+1) = 2 z where z is a natural no.  be true
  k² + k =2 z 
  k² = 2 z - k               - 1
step 3 p(k+1) = k+1(k +1 +1) is true we have to prove this
  (k+1)( k+2) = k² + 3 k + 2 
2 z  - k +3 k +2 = 2 z + 2 k + 2          - by 1 
2[ z +k + 1]           it will be a even no.
which means that p(k+1) is true  
hope this will help you