Proving Divisibility by Induction

[Jamin2112]

Here are some that I'm stuck on.

Pg. 56, #12

Prove by induction on n that, for all positive integers n, 3 divides 4^n + 5

Of course, the base case it is P(1) = (4^1 + 5) / 3 = 9/3 = 3...TRUE!

I just can't see the trick here. P(K+1)= (4^(K+1) + 5) / 3 = ((4)(4^K) + 5)/3= ... not getting anywhere, really.

Pg. 55, #17

For a positive integer n the number An is defined by
A1=1 [supposed to A with a subscript 1]
Ak+1 [supposed to A with a subscript k+1] = (6Ak+5)/(Ak+2)

Prove by induction on n that, for all positive integers (i) An>0 and (ii) An<5

I see by long division that we have
Ak+1=1-7/(Ak+2)... not sure if that helps. I know that Ak+1 will be zero with Ak=5...no sure if that helps though...

 

[economist13]

#12) Consider adding 0 in a creative way...so that you can factor out a 4 completely and have the sum of two numbers that are both divisible by 3.