Answer :

To Prove:


1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1)


Steps to prove by mathematical induction:


Let P(n) be a statement involving the natural number n such that


(i) P(1) is true


(ii) P(k + 1) is true, whenever P(k) is true


Then P(n) is true for all n ϵ N


Therefore,


Let P(n): 1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1)


Step 1:


P(1) = 1/2 1(1 + 1) = 1/2 × 2 = 1


Therefore, P(1) is true


Step 2:


Let P(k) is true Then,


P(k): 1 + 2 + 3 + 4 + … + k = 1/2 k(k + 1)


Now,


1 + 2 + 3 + 4 + … + k + (k + 1) = 1/2 k(k + 1) + (k + 1)


= (k + 1){ 1/2 k + 1}


= 1/2 (k + 1) (k + 2)


= P(k + 1)


Hence, P(k + 1) is true whenever P(k) is true


Hence, by the principle of mathematical induction, we have


1 + 2 + 3 + 4 + … + n = 1/2 n(n + 1) for all n ϵ N


Hence proved.


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