how to prove that this is true for all natural numbers n, using principle of mathematical induction.
(3i+2)= 5+8+11+14+...+ (3n+2)=3n^2+7n/2
The easy way. You have surely seen the proof by induction that
n
∑ k = n(n+1)/2
k=1
This sum is just
n
∑ 3k+2 = 3∑k + ∑2 = 3n(n+1)/2 + 2n = (3n^2+7n)/2
k=1
The hard way:
k=1: 5 = 3*1+2 so P(1)
Assume P(n). Now, if k=n+1,
5+8+...+3n+2+3(n+1)+2 = (3n^2+7n)/2 + 3(n+1)+2
= (3n^2 + 7n + 6n+10)/2
= (3(n^2+2n+1)+7n+7)/2
= (3(n+1)^2 + 7(n+1))/2
= P(k+1)