# Using mathematical induction prove that the sum of the first n integers is n*(n+1)/2.

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### 2 Answers

Let us express the formula as: 1+2+3…n = n*(n+1)/2

As we are supposed to prove it using mathematical induction, we have to prove that it is true for n = 1 and that if it is true for n we can show that it is true for n+1.

First for n=1. The sum = 1*2/2=1 . So it is true.

Now given that 1+2+3..n= n*(n+1)/2 we have to show that 1+2+3…(n+1)= (n+1)*(n+2)/2

If 1+2+3..n= n*(n+1)/2

=> 1+2+3…(n+1) =[ n*(n+1)/2] + n+1

= (n^2 +n)/2 + 2(n+1)/2

= (n^2 +n +2n +2 )/2

=(n^2 + 2n +n+2)/2

=[n(n+2) +1(n+2)]/2

=(n+1)*(n+2)/2

So we prove that the expression is true for n+1 given that it is true for n.

**Hence by mathematical induction the formula for the sum of the first n numbers is given as n*(n+1)/2 has been proved.**

We presume the formula Sn = n(n+1)/2 is true for the sum of first n natural numbers only.

Now we use the same formula to find the sum of the first n+1 natural numbers.

Then Sn+1 = Sn + (n+1) = n(n+1)/2 +(n+1).

Sn+1 = n(n+1)/2 + 2(n+1)/2

Sn+1 = {(n+1)/2} {n+2}

Sn+1 = (n+1)(n+2)/2

Sn+1 = (n+1)[(n+1)+1]/2. So it is as good as substituting n+1 in place of n in the formula Sn = n(n+1)/2.

Therefore if Sn = n(n+1)/2 is true for n , then Sn+1 = (n+1)(n+2)/2 is also true for n+1.

Now we take S1 = 1 obviously. S1 = 1(1+1)/2 = 1 is true by formula.

So S1 =1(1+1)/2 is true. S2 = 2(2+1)/2 true by induction. So Sn = n(n+1)/2 is true for all n.