Prove the following by using principle of mathematical ∀n ∈ M 11n

1.	Prove the following by using principle of mathematical ∀n ∈ M 11n+2 + 112n+1 is divisible by 133.
Answer» Let P(n) ∶ 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133. Step 1: P(1) ∶ 11¹⁺² + 12^{2 ×1+1} = 1331 + 1728 = 3059 \(\frac{3059}{133}\) = 23, hence, it is true for n = 1. Step 2: Let it be true for n = k, i.e., 11^k+2 + 12^2k+1 is divisible by 133. ⇒ P(k) ∶ 11^k+2 + 122^k+1 = 133. λ For some λ ∈ M. ...(i) Step 3: To prove that P(k + 1) = 11^k+3 + 12^2(k+1)+1 is also divisible by 133. Now, 11^k+3 + 12^2k+3 = 11.11^k+2 + 12^2k+3 = 11.(133λ − 12^2k+1 ) + 12^2k+3 ...[From (i)] = 11.133λ − 11.12^2k+1 + 12^2k+3 = 11.133λ − 11.12^2k+1 + 12².12^2k+1 = 11.133λ + 12^2k+1 (144 − 11) = 11.133λ + 12^2k+1.133 = 133(11λ + 12^2k+1), Which is divisible by 133 for some λ ∈ N. ∴ P(k + 1) is true. Hence, by Principle of mathematical induction P(n) is true for all n ∈ N.

Prove the following by using principle of mathematical ∀n ∈ M 11n+2 + 112n+1 is divisible by 133.

Answer»

Let P(n) ∶ 11ⁿ⁺² + 12²ⁿ⁺¹ is divisible by 133.

Step 1: P(1) ∶ 11¹⁺² + 12^{2 ×1+1} = 1331 + 1728 = 3059

\(\frac{3059}{133}\) = 23, hence, it is true for n = 1.

Step 2: Let it be true for n = k, i.e.,

11^k+2 + 12^2k+1 is divisible by 133.

⇒ P(k) ∶ 11^k+2 + 122^k+1 = 133.

λ For some λ ∈ M. ...(i)

Step 3: To prove that

P(k + 1) = 11^k+3 + 12^2(k+1)+1 is also divisible by 133.

Now, 11^k+3 + 12^2k+3

= 11.11^k+2 + 12^2k+3

= 11.(133λ − 12^2k+1 ) + 12^2k+3 ...[From (i)]

= 11.133λ − 11.12^2k+1 + 12^2k+3

= 11.133λ − 11.12^2k+1 + 12².12^2k+1

= 11.133λ + 12^2k+1 (144 − 11)

= 11.133λ + 12^2k+1.133

= 133(11λ + 12^2k+1),

Which is divisible by 133 for some λ ∈ N.

∴ P(k + 1) is true.

Hence, by Principle of mathematical induction P(n) is true for all n ∈ N.