Use the method of mathematical induction to prove that \({4^n} + 15n - 1\) is divisible by 9 for \(n \in {\mathbb{Z}^ + }\).

let \(P(n)\) be the proposition that \({4^n} + 15n - 1\) is divisible by 9

showing true for \(n = 1\)     A1

ie\(\,\,\,\,\,\)for \(n = 1,{\text{ }}{4^1} + 15 \times 1 - 1 = 18\)

which is divisible by 9, therefore \(P(1)\) is true

assume \(P(k)\) is true so \({4^k} + 15k - 1 = 9A,{\text{ }}(A \in {\mathbb{Z}^ + })\)     M1

Note:     Only award M1 if “truth assumed” or equivalent.

consider \({4^{k + 1}} + 15(k + 1) - 1\)

\( = 4 \times {4^k} + 15k + 14\)

\( = 4(9A - 15k + 1) + 15k + 14\)     M1

\( = 4 \times 9A - 45k + 18\)     A1

\( = 9(4A - 5k + 2)\) which is divisible by 9     R1

Note:     Award R1 for either the expression or the statement above.

since \(P(1)\) is true and \(P(k)\) true implies \(P(k + 1)\) is true, therefore (by the principle of mathematical induction) \(P(n)\) is true for \(n \in {\mathbb{Z}^ + }\)     R1

Note:     Only award the final R1 if the 2 M1s have been awarded.

[6 marks]