17M.1.AHL.TZ1.H_8

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

showing true for $n=1$     A1

ie$$for $n=1,4^1+15\times 1-1=18$

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

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

[N/A]