Prove that for every positive integer n ≥ 1, n2+n is even integer using mathematical induction.

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Suresh Chand · Accepted Answer

Let p(n) = n2 + n First we perform basic steps We show p(1) is true i.e. p(1) = 12 + 1 = 2 (which is divisible by 2) Hence, n2 + 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. k2 + 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) = k2 + 2k + 1 + k  + 1 = k2 + 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, n2 + n is even whenever p ≥ 1

Prove that for every positive integer n ≥ 1, n²+n is even integer using mathematical induction.

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Prove that for every positive integer n ≥ 1, n2+n is even integer using mathematical induction.

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Prove that for every positive integer n ≥ 1, n²+n is even integer using mathematical induction.