We want to prove that P(n)= 1+2+3++(n-1)+n = n(n+1) 2 , for all n.

We start by confirming the equation for the case n=1.
For P(1), we have P(1)=1 =1×22

Then assume the formula is true for the kth case:
That is, P(k)= 1+2+3++(k-1)+k = k(k+1) 2

For the (k+1)th case, we have
P(k+1) = 1+2+3++(k-1)+k+(k+1) = P(k)+(k+1) = k(k+1) 2 +(k+1) = k(k+1) 2 + 2(k+1) 2 = (k+1)(k+2) 2