Use induction to prove that 10n + 3 × 4n + 2 + 5, is divisible by 9,

1.	Use induction to prove that 10n + 3 × 4n + 2 + 5, is divisible by 9, for all natural numbers n.
Answer» P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9 P(1) = 10¹ + 3 × 4² + 5 = 10 + 3 × 16 + 5 = 10 + 48 + 5 = 63 ÷ by 9 So P(1) is true. Assume that P(k) is true (i.e.) 10^k + 3 × 4^{k + 2} + 5 is ÷ by 9 (i.e.) 10^k + 3 × 4^{k + 2} + 5 = 9C (where C is an integer) ⇒ 10^k = 9C – 5 – 3 × 4^{k + 2} ….. (1) To prove P(k + 1) is true. Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5 = 10 × 10^k + 3 × 4 × 4^{k + 2} + 5 = 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5 = 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5 = 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5 = 90C – 45 – 18 × 4^{k + 2} = 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9 So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.

Use induction to prove that 10n + 3 × 4n + 2 + 5, is divisible by 9, for all natural numbers n.

Answer»

P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9

P(1) = 10¹ + 3 × 4² + 5 = 10 + 3 × 16 + 5

= 10 + 48 + 5 = 63 ÷ by 9

So P(1) is true. Assume that P(k) is true

(i.e.) 10^k + 3 × 4^{k + 2} + 5 is ÷ by 9

(i.e.) 10^k + 3 × 4^{k + 2} + 5 = 9C (where C is an integer)

⇒ 10^k = 9C – 5 – 3 × 4^{k + 2} ….. (1)

To prove P(k + 1) is true.

Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5

= 10 × 10^k + 3 × 4 × 4^{k + 2} + 5

= 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5

= 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5

= 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5

= 90C – 45 – 18 × 4^{k + 2}

= 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9

So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.