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