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Login NowLet p(n) = n^{2} + n

First we perform basic steps

We show p(1) is true

i.e. p(1) = 1^{2} + 1 = 2 (which is divisible by 2)

Hence, n^{2} + n is an even integer when n = 1

**Induction Step:**

Let us suppose, for any arbitary k, p(k) is true. i,e, it is divisible by 2 so that it is an even integer.

i.e. k^{2} + k = 2t, where t is an integer

Now, we have to show p(k + 1) is also even i.e. divisible by 2

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

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

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

= 2t + 2k + 2

= 2t + 2(k+1)

= 2t + 2m (where m is an integer)

= 2(t+m)

= 2s (where is an integer)

Here, 2s is divisible by 2. Som p(x + 1) is an even integer. So, its true.

Hence, by matematical induction, n^{2} + n is even whenever p ≥ 1

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