Mathematical induction 3

Question

For the given statement P_n, write the statements P₁, P_k, and P_k+1.

 

2 + 4 + 6 + . . . + 2n = n(n+1) 

Answer 1

Statement : P_n = 2 + 4 + 6 + . . . . . + 2n = n(n + 1).

• Check the statement whether it is true for n = 1.

Left hand side expression = 2n = 2(1) = 2

Right hand side expression = n(n + 1) = 1(1+1) = 2

Therefore, for n = 1 the statement (P₁) is true.  ---------> part 1

 

Lets assume n = k, then the statement

P_k = 2 + 4 + 6 + . . . . . + 2k  = k(k + 1)  is true.  ---------> equation 1

 

• So the statement must true for n = k + 1.

P_k+1 = 2 + 4 + 6 + . . . . . + 2k + 2(k+1)

         = [2 + 4 + 6 + . . . . . + 2k ] + 2k + 2

         = P_k + 2(k + 1)

         = k(k + 1)+ 2(k + 1)

         = k(k + 1) + 2(k + 1)

P_k+1 = (k+1)*[(k +1)+ 1]    -------> part 2

P_k+1 = 2 + 4 + 6 + . . . . . + 2(k+1) = (k+1)*[(k +1)+ 1]

From results part 1 and 2, we can conclude by mathematical induction that the formula is valid for all positive integers values of n.

The values of P₁ = 2, P_k = k(k + 1) and P_k+1 = (k+1)*(k +2).