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

1.	If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 1 + 2 + 22 + … + 2n = 2n + 1 – 1 for all n ϵ N
Answer» Let P(n) = 1 + 2 + 2² + … + P(n): 1 + 2 + 2² + … + 2ⁿ= 2^{n + 1} – 1 for all n ϵ N Step1: P(1) = 1 = (2) – 1 = 1 Thus, P(n) is equal to 2 n + 1 – 1 for n = 1 Step2: Let, P(m) be equal to 2^{m + 1} – 1 Then, 1 + 2 + 2² + … + 2^m = 2^{m + 1} – 1 Now, we need to show that P(m+1) is true whenever P(m) is true. P(m+1) = 1 + 2 + 2² + … + 2^m + 2^{m + 1} = 2^{m + 1}– 1 + 2^{m + 1} = 2.2^{m + 1} –1 = 2^{m + 2} – 1 Thus, P(m+1) is true. So, by the principle of mathematical induction, P(n) is true for all nϵN.

If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 1 + 2 + 22 + … + 2n = 2n + 1 – 1 for all n ϵ N

Answer»

Let P(n) = 1 + 2 + 2² + … +

P(n): 1 + 2 + 2² + … + 2ⁿ= 2^{n + 1} – 1 for all n ϵ N

Step1:

P(1) = 1 = (2) – 1 = 1

Thus, P(n) is equal to 2 n + 1 – 1 for n = 1

Step2:

Let, P(m) be equal to 2^{m + 1} – 1

Then, 1 + 2 + 2² + … + 2^m = 2^{m + 1} – 1

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

P(m+1) = 1 + 2 + 2² + … + 2^m + 2^{m + 1}

= 2^{m + 1}– 1 + 2^{m + 1}

= 2.2^{m + 1} –1

= 2^{m + 2} – 1

Thus, P(m+1) is true.

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