If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 2

1.	If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 2.7n + 3.5n – 5 is divisible by 24 for all n ϵ N
Answer» Let P(n) = 2.7ⁿ+ 3.5ⁿ– 5 Now, P(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n ϵ N Step1: P(1) = 2.7 + 3.5 – 5 = 1.2 Thus, P(1) is divisible by 24 Step2: Let, P(m) be divisible by 24 Then, 2.7^m + 3.5^m – 5 = 24λ, where λ ϵ N. Now, we need to show that P(m+1) is true whenever P(m) is true. So, P(m+1) = 2.7^m+1 + 3.5^m+1 – 5 = 2.7^m+1 + 5.( 2.7^m + 3.5^m – 5 ) – 5 = 2.7^m+1 + 5.( 24λ + 5 - 2.7^m ) – 5 = 2.7^m+1 + 120λ + 25 - 10.7^m – 5 = 2.7^m.7 - 10.7^m+ 120 λ +24 – 4 = 7^m(14 – 10) + 120 λ +24 – 4 = 7^m(4) + 120 λ +24 – 4 = 7^m(4) + 120 λ +24 – 4 = 4(7^m - 1) + 24(5λ +1) As, 7^m – 1 is a multiple of 6 for all m ϵ N. So, P(m+1) = 4.6μ +24(5λ +1) = 24(μ +5λ +1) 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)? 2.7n + 3.5n – 5 is divisible by 24 for all n ϵ N

Answer»

Let P(n) = 2.7ⁿ+ 3.5ⁿ– 5

Now, P(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n ϵ N

Step1:

P(1) = 2.7 + 3.5 – 5 = 1.2

Thus, P(1) is divisible by 24

Step2:

Let, P(m) be divisible by 24

Then, 2.7^m + 3.5^m – 5 = 24λ, where λ ϵ N.

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

So, P(m+1) = 2.7^m+1 + 3.5^m+1 – 5

= 2.7^m+1 + 5.( 2.7^m + 3.5^m – 5 ) – 5

= 2.7^m+1 + 5.( 24λ + 5 - 2.7^m ) – 5

= 2.7^m+1 + 120λ + 25 - 10.7^m – 5

= 2.7^m.7 - 10.7^m+ 120 λ +24 – 4

= 7^m(14 – 10) + 120 λ +24 – 4

= 7^m(4) + 120 λ +24 – 4

= 4(7^m - 1) + 24(5λ +1)

As, 7^m – 1 is a multiple of 6 for all m ϵ N.

So, P(m+1) = 4.6μ +24(5λ +1)

= 24(μ +5λ +1)

Thus, P(m+1) is true.

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