Use iteration to guess an explicit formula for the sequence and use mathematical induction to verify the formula obtained:

Question

Pk = Pk-1 + 2 * 3k, for each integer K ≥ 2, where P_k = 2

Answer 1

Assuming that P₁=2, then

P₂ = P₁+2*3(2) = 2+2*3(2). = 2*[1+3*2]

P₃ = P₂+2*3(3) = 2+2*3(2)+2*3(3) = 2*[1+3*2+3*3]

P₄ = P₃ + 2*3(4) = 2+2*3(2)+2*3(3)+2*3(4) = 2*[1+3*2+3*3+3*4]

Continuing this trend, we find that for k>=2 that

P_k = 2*[1+3*2+3*3+3*4+...+ 3*k] = 2*[-2+3*1+3*2+3*3+3*4+...+3*k] = 2*[-2+3[1+2+...+k]] = 2*[-2+ 3k(k+1)/2] = 3k²+3k-4

(Note that the part in bold was rewritten so that the pattern started with the term 3*1; furthemore, we use the fact that 1+2+...+k = k(k+1)/2 for any k>=1).

We now verify this formula using mathematical induction.

We first justify the base case, which is k=1 (the initial condition). We find that

P₁=3*1²+3*1-4=6-4=2, which agrees with the intial value given for P₁.

Next, we state the inductive hypothesis: For any k>=1, we have that P_k = 3k²+3k-4.

Finally, we perform the inductive step; if the statement is true for arbitrary k, then it should be true for k+1. Specifically, we need to conclude that P_k+1=3(k+1)²+3(k+1)-4.

From the iterative definition, we know that P_k+1=P_k+2*3(k+1). By the inductive hypothesis, we know that P_k = 3k²+3k-4. Hence, we have that

P_k+1 = 3k²+3k-4 + 2*3(k+1) = 3k(k+1)-4+2*3(k+1) = 3(k+1)(k+2) - 4 = 3(k+1)([k+1]+1) - 4 = 3(k+1)²+3(k+1)-4.

This completes the inductive step, and thus we have proved that the explicit formula for

P_k=P_k-1+2*3k (k>=2) with P₁=2

is given by P_k = 3k²+3k-4 for any k>=1.