Question

Use induction to prove that 5ⁿ⁺¹ + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n

Answer 1

P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20

P(1) = 5¹⁺¹ + 4 × 6¹ – 9

= 5² + 24 – 9

= 25 + 24 – 9

= 40 ÷ by 20

So P(1) is true

Assume that the given statement is true for n = k

(i.e) 5^{k + 1} + 4 × 6ⁿ – 9 is ÷ by 20

P(1) = 5¹⁺¹ + 4 × 6¹ – 9

= 25 + 24 – 9

So P(1) is true

To prove P(k + 1) is true

P(k + 1) = `5^("k" + 1 + 1) + 4 × 6^("k" + 1 + 1) – 9`

= 5 × 5 ^{k + 1} + 4 × 6 × 6^k – 9

= 5[20C + 9 – 4 × 6^k] + 24 × 6^k – 9  .....[from(1)]

= 100C + 45 – 206^k + 246^k – 9

= 100C + 46^k + 36

= 100C + 4(9 + 6^k)

Now for k = 1

⇒ 4(9 + 6^k)

= 4(9 + 6)

= 4 × 15 = 60 ÷ by 20 .

For k = 2

= 4(9 + 6²)

= 4 × 45

= 180 ÷ 20

So by the principle of mathematical induction 4(9 + 6^k) is ÷ by 20

Now 100C is ÷ by 20.

So 100C + 4(9 + 6^k) is ÷ by 20

⇒ P(k + 1) is true

Whenever P(k) is true.

So by the principle of mathematical induction

P(n) is true.