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

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.