Prove that n^{2} + n is even , where n is natural number.?

step 1. put n=1

n^{2 }+ n = 1^{2} +1 =2 which is even

step 2 . let assume result is true for n=k

then k^{2} +k is even number

step 3. to prove for n=k+1

(k+1)^{2} + (k+1)

= k^{2} + 1 + 2k + k + 1

= k^{2} + k + (2k +2)

i.e = even number + 2 (k+1) [using 1 . & 2(k+1) is also even ]

(also an even for all n belomgs to N)

therefore result is true for for all n belongs to N

hope it helps u