using induction, prove that 4 divides 3^(2n+1)+5 for every integer when n is greater than or equal zero?

any help is appreciated! thanks!

step by step would be great.

4 Answers

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  • 8 years ago

    Hello,

    Does 4 divide (5 + 3²ⁿ⁺¹) for any natural integer n ?

    ♥ Let's check with n=0:

    When n=0,

    5 + 3²ⁿ⁺¹ = 5 + 3¹ = 8

    which is indeed divisible by 4.

    ♥ Let's suppose it's true for rank n:

    (5 + 3²ⁿ⁺¹) is divisible by 4.

    ♥ Let's prove it for rank n+1:

    5 + 3²ⁿ⁺³ = 5 + 3²ⁿ⁺¹ × 3²

       = 5 + (5 + 3²ⁿ⁺¹ - 5) × 9

       = 5 + 9(5 + 3²ⁿ⁺¹) - 9×5

       = 9(5 + 3²ⁿ⁺¹) - 45 + 5

       = 9(5 + 3²ⁿ⁺¹) - 40

    Since (5 + 3²ⁿ⁺¹) is divisible by 4, 9(5 + 3²ⁿ⁺¹) is obviously also divisible by 4.

    Since 40 is divisible by 4.

    Thus (5 + 3²ⁿ⁺³) is the difference of two values that are divisible by for 4.

    Hence (5 + 3²ⁿ⁺³) is divisible by 4.

    Thus is the property proven at rank n+1.

    ♥ We demonstrated by mathematical induction that

    4 divide (5 + 3²ⁿ⁺¹) for any natural integer n.

    (In fact, the same induction can prove it's divisible by 8.)

    Logically,

    Dragon.Jade :-)

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  • Anonymous
    8 years ago

    Assume that the given expression is divisible by 4 (i.e., P(n) is true). Then we need to show that implies the expression w/ n -> n + 1 is also divisible by 4 (i.e., show P(n) => P(n + 1)...)

    Let's replace n -> (n + 1) in our given expression:

    (3 ^ (2(n + 1) + 1)) + 5 can be rewritten as 3 ^ (2n + 3) + 5

    ==> 3 ^ [(2n + 1) + 2] + 5

    BUT: 3 ^ [(2n + 1) + 2] = (3 ^ (2n + 1)) * 3^2, which becomes 9(3 ^ (2n + 1))

    Substituting this in, we arrive at:

    ==> 9(3 ^ (2n + 1)) + 5 for expression. We can break this apart to form

    ==> 8(3 ^ (2n + 1)) + 3 ^ (2n + 1) + 5 ==> 8(3 ^ (2n + 1)) + [3 ^ (2n + 1) + 5]

    However, note that we assumed the expression in brackets (which is P(n)) was divisible by 4 (at the start), and obviously 8(3 ^ (2n + 1)) is divisible by 4, so the whole expression must also be divisible by 4.

    In other words P(n) => P(n + 1) for our induction hypothesis.

    Now just go and check that P(1) works (i.e., that the expression is divisible by 4 when n = 1), and that should give you everything you need!

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  • 8 years ago

    go to cramster and ask them. Its 15 a month.

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  • ingles
    Lv 4
    3 years ago

    5 = a million (mod 4) so then 5^n = a million^n (mod 4) = a million (mod 4) And so gazing 5^n - a million (mod 4) we get 5^n - a million (mod 4) = a million^n - a million (mod 4) = a million-a million (mod 4) = 0 (mod 4). subsequently 4 divides 5^n - a million

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