If n is an integer, n \(\geq\)1, then show that \(3^{2^n}\)– 1 is di

1.	If n is an integer, n \(\geq\)1, then show that \(3^{2^n}\)– 1 is divisible by 2n + 2.
Answer» Let T(n) be the statement: \(3^{2^n}\)– 1 is divisible by 2^{n + 2} Basic Step: For n = 1, \(3^{2^1}\) – 1 = 8 and 2^{n + 2} = 8 ⇒ T (1) is true Induction Step: Assume T(k) to be true, i.e., T(k) = \(3^{2^k}\) – 1 is divisible by 2^{k + 2} = \(3^{2^k}\) – 1 = m. 2^{k + 2} when m∈N ...(i) = \(3^{2^k}\)= m. 2^{k + 2} + 1 Now we need to prove that T(k + 1) holds true. ∴ \(3^{2^{k+1}}\) –1 = \(3^{2^k.2}\) – 1 = (m . 2^{k + 2} + 1)² – 1 (using (i)) = m² (2^k+2)²+ 2m . 2^k+2 + 1 – 1 = 2^k+2 (m² . 2^k+2 + 2m) ⇒ T(k + 1) = \(3^{2^{k+1}}\) – 1 is divisible by 2^k+2, whenever T(k) holds. Thus \(3^{2^n}\) – 1 is divisible by 2^{n + 2} for all integers n \(\geq\) 1.

If n is an integer, n \(\geq\)1, then show that \(3^{2^n}\)– 1 is divisible by 2n + 2.

Answer»

Let T(n) be the statement:

\(3^{2^n}\)– 1 is divisible by 2^{n + 2}

Basic Step:

For n = 1,

\(3^{2^1}\) – 1 = 8 and 2^{n + 2} = 8

⇒ T (1) is true

Induction Step:

Assume T(k) to be true, i.e.,

T(k) = \(3^{2^k}\) – 1 is divisible by 2^{k + 2}

= \(3^{2^k}\) – 1 = m. 2^{k + 2} when m∈N ...(i)

= \(3^{2^k}\)= m. 2^{k + 2} + 1

Now we need to prove that T(k + 1) holds true.

∴ \(3^{2^{k+1}}\) –1 = \(3^{2^k.2}\) – 1

= (m . 2^{k + 2} + 1)² – 1 (using (i))

= m² (2^k+2)²+ 2m . 2^k+2 + 1 – 1 = 2^k+2 (m² . 2^k+2 + 2m)

⇒ T(k + 1) = \(3^{2^{k+1}}\) – 1 is divisible by 2^k+2, whenever T(k) holds.

Thus \(3^{2^n}\) – 1 is divisible by 2^{n + 2} for all integers n \(\geq\) 1.