Use mathematical induction to prove the statement is true for all positive integers n.1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + n(n + 1) =

Question

What will be an ideal response?

Answer 1

Answers may vary. Possible answer:
First, we show the statement is true when n = 1.
For n = 1, we get 1 ? 2 =   

 Since  =  = 1 ? 2, P₁ is true and the first condition for the principle of induction is satisfied.
Next, we assume the statement holds for some unspecified natural number k. That is,
P_k: 1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + k(k + 1) =   is assumed true.
On the basis of the assumption that P_k is true, we need to show that P_k+1 is true. 
P_k+1: 1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + k(k + 1) + (k + 1)(k + 2) = 
So we assume that  is true and add the next term,  to both sides of the equation.
1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + k(k + 1) + (k + 1)(k + 2) =  + (k + 1)(k + 2)
 1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + k(k + 1) + (k + 1)(k + 2) =  
 1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + k(k + 1) + (k + 1)(k + 2) =  
The last equation says that P_k+1 is true if P_k is assumed to be true. Therefore, by the principle of mathematical induction, the statement 1 ? 2 + 2 ? 3 + 3 ? 4 + . . . + n(n + 1) =  is true for all natural numbers n.