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Homework Help: A few proofs by induction

  1. Jan 20, 2010 #1
    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......
  2. jcsd
  3. Jan 20, 2010 #2
    #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.
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