Prove by induction?

1. Dec 16, 2007

pauliedangerous

can anyone help me with this - my notes make this slightly confusing and struggling to finsih it.

prove by induction that fo0r every positive integer (n) that 13n - 6n is divisible by 7

ive got so far:

(n = 1) = 13(1) - 6(1) = 7 which is divisible by 7

n(k+1) = 13(k+1) - 6(k+1)
= 13(1).13(k) - 6(1).6(k)
= 13.13k - 6.6k

then stuck as my notes with examples give calculations tht i dont usnderstand where they come from il post them if you can explain it.

example from my notes:

Question 8: Where n is an integer prove that the expression 7n – 2n is exactly divisible by 5.
Hint: 7k+1 – 2k+1 can be written as 7k.71 – 2k.21 i.e. 7k.7 – 2k.2

Question 8: It holds for P(1) = 71 – 21 = 5 which is exactly divisible by 5.
Inductive step: Consider P(k+1) assuming P(k) is true.
P(k +1) = 7k+1 – 2k+1
= 71.7k - 21.2k
= 7.7k - 2.2k
= 7k + 6.7k - 2k - 2k splitting up the terms 7.7k and 2k
= (7k - 2k) + 6.7k - 2k and (7k - 2k) = P(k)
= P(k) + 6.7k - 2k
= P(k) + 5.7k + 7k - 2k splitting up the term 6.7k
= P(k) + 5.7k + P(k) 7k - 2k is P(k)
This expression and thus P(k+1) is exactly divisible by 5 since both expressions P(k) and 5.7k are exactly divisible by 5.

P(1) is true and P(k)→P(k+1), so by the principle of induction P(n) holds for all n.

thanks for any help

2. Dec 16, 2007

EnumaElish

You have:
p(k+1) = 13(k+1) - 6(k+1)
= 13(1).13(k) - 6(1).6(k)
= 13.13k - 6.6k

Note 13x = 12x + x and 6y = 5y + y. Let x = 13k and y = 6k.

You can continue as:
p(k+1) = 12 13k + 13k - (5 6k + 6k) = 12 13k - 5 6k + (13k - 6k) = 156k - 30k + p(k)
...

You should pay attention to your parentheses, e.g. 7k + 1 is different from 7(k+1).