If P(n) is the statement “n(n + 1) is even”, then what is P(3)? n

1.	If P(n) is the statement “n(n + 1) is even”, then what is P(3)? n3 – 7n + 3 is divisible by 3 for all n ϵ N.
Answer» Let P(n) = n³ – 7n + 3 Now, P(n): n³ – 7n + 3 is divisible by 3 for all n ϵ N Step1: P(1) = 1 – 7 + 3 = -3 Thus, P(1) is divisible by 3 Step2: Let, P(m) be divisible by 24 Then, n³ – 7n + 3 = 3λ, where λ ϵ N. Now, we need to show that P(m+1) is true whenever P(m) is true. So, P(m+1) = (n+1)³ – 7(n+1) + 3 = n³+3n²+3n+1-7n-7+3 = n³– 7n + 3 +3n²+3n+1-7 = 3λ+3(n²+n-2) = 3(λ+n²+n-2) Thus, P(m+1) is true. So, by the principle of mathematical induction, P(n) is true for all n ϵN.

If P(n) is the statement “n(n + 1) is even”, then what is P(3)? n3 – 7n + 3 is divisible by 3 for all n ϵ N.

Answer»

Let P(n) = n³ – 7n + 3

Now, P(n): n³ – 7n + 3 is divisible by 3 for all n ϵ N

Step1:

P(1) = 1 – 7 + 3 = -3

Thus, P(1) is divisible by 3

Step2:

Let, P(m) be divisible by 24

Then, n³ – 7n + 3 = 3λ, where λ ϵ N.

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = (n+1)³ – 7(n+1) + 3

= n³+3n²+3n+1-7n-7+3

= n³– 7n + 3 +3n²+3n+1-7

= 3λ+3(n²+n-2)

= 3(λ+n²+n-2)

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all n ϵN.