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) = 4k + 15k - 1 = 9A
Then it should be true that for n = (k + 1)
If P(k+1) = 4k+1 + 15k + 15 - 1= 9B
==> 3(4)k + 15 + (4k + 15k - 1)
==> 3(4)k + 15 + 9A
==> 3(4k + 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.
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.
+33614 P(n) = 4n + 15n - 1. P(1) = 18. Divisible by 9
P(n+1) = P(n) + Q(n). (A) where Q(n) = 3*4n + 15
Q(n) = 3*4n + 15. Q(1) = 27. Divisible by 9
Q(n+1) = 3*4n+1 + 15. Q(n+1) = 12*4n + 15 Q(n+1) = 9*4n + 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.
