Use induction to prove that 5n + 1 + 4 × 6n when divided by 20 leaves

1.	Use induction to prove that 5n + 1 + 4 × 6n when divided by 20 leaves a remainder 9, for all natural numbers n.
Answer» P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20 P(1) = 5^{1 + 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) = 5^{1 - 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.

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

Answer»

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

P(1) = 5^{1 + 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) = 5^{1 - 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.