Question: Use the method of mathematical induction to prove that 4^n + 15n -1 is divisible by 9. n is a positive integer. (OK!)

True for n = 1

P(1) ......... = 9A

True for n = k

If P(k) = 4^{k} + 15k - 1 = 9A

Then it should be true that for n = (k + 1)

If P(k+1) = 4^{k+1} + 15k + 15 - 1= 9B

==> 3(4)^{k} + 15 + (4^{k} + 15k - 1)

==> 3(4)^{k} + 15 + 9A

==> 3(4^{k }+ 5 + 3A)

3 =/= 9

6 marks question, why did I derived 3 instead of 9. Someone help? HL Maths IB. I hate it when I follow steps and then end up backed in a corner.

Guest Oct 13, 2017

#1**0 **

There could be something obvious that I'm not seeing either, but failing that, a second induction proof to show that 4^k + 5 is divisible by 3 works.

Guest Oct 13, 2017

#2**+1 **

P(n) = 4^{n} + 15n - 1. P(1) = 18. Divisible by 9

P(n+1) = P(n) + Q(n). (A) where Q(n) = 3*4^{n} + 15

Q(n) = 3*4^{n} + 15. Q(1) = 27. Divisible by 9

Q(n+1) = 3*4^{n+1} + 15. Q(n+1) = 12*4^{n} + 15 Q(n+1) = 9*4^{n} + Q(n).

Hence if Q(n) divisible by 9 then Q(n+1) divisible by 9

Hence Q(n) divisible by 9 (because Q(1) also divisible by 9)

So if P(n) divisible by 9 then P(n+1) divisible by 9 (because both terms on RHS of A are divisible by 9).

Since P(1) is also divisible by 9 then P(n) is divisible by 9. QED.

Alan Oct 13, 2017