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Math Help - solve that

  1. #1
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    solve that

    1) show that 9/ (n^3+(n+1)^3+(n+2)^3) ,n belongs N ,throhg
    a) induction b)direct

    2)show that

    57/(7^n+2 +8^2n+1) , n blongs N , throgh

    a) induction b)congrunce
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  2. #2
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    Hello, hoger!

    Here's the first one . . .


    1) Show that \bigg[n^3+(n+1)^3+(n+2)^3\bigg] is a multiple of 9, for n \in N

    a) by induction
    S(n)\!:\;\;n^3 + (n+1)^3 + (n+2)^3 is a multiple of 9.


    Verify S(1)\!:\;\;1^3+2^3+3^3 \:=\:36, which is a multiple of 9.


    Assume S(k)\!:\;\;k^3+(k+1)^3  + (k+2)^3 \:=\:9a, for some integer a.


    Add (k+3)^3 - k^3 to both sides:

    . . k^3 +(k+1)^3 + (k+2)^3 + {\color{blue}(k+3)^3 - k^3} \;=\;9a + {\color{blue}(k+3)^3 - k^3}

    . . \underbrace{(k+1)^3 + (k+2)^3 + (k+3)^3}_{\text{the left side of }S(k+1)} \;=\;9a + (k+3)^3 - k^3

    We must show that the right side is a multiple of 9.


    The right side is: . 9a + k^3 + 9k^2 + 27k + 27 - k^3

    . . = \;9a + 9k^2 + 27k + 27 \;=\;\underbrace{9(a + k^2 + 3k + 3)}_{\text{a multiple of 9}}


    The inductive proof is complete.




    b) by direct proof.
    N \;=\;n^3 + (n+1)^3 + (n+2)^3 \;=\;n^3 + (n^3 + 3n^2 + 3n + 1) + (n^3 + 6n^2 + 12n + 8)

    . . = \;3n^3 + 9n^2 + 15n + 9  \;=\;3(n^3 + 3n^2 + 5n + 3) \;=\;3(n+1)(n^2+2n+3)

    We must show that either (n+1) or (n^2+2n+3) is a multiple of 3.


    There are three cases for n

    . . \begin{array}{ccc}(1) & n\text{ is one less than a multiple of 3:} & n \:=\:3a - 1\\<br />
(2) &n\text{ is a multiple of 3:} & n\:=\:3a \\<br />
(3) &n\text{ is one more than a multiple of 3:} & n \:=\:3a+1 \end{array}


    (1)\;\;n \:=\:3a-1
    . . .Then: . n + 1 \:=\:(3a-1) + 1 \:=\:3a, a multiple of 3.

    (2)\;\;n \:=\:3n
    . . . \text{Then: }\;n^2+2n+3 \;=\;(3a)^2 + 2(3a) + 3 \:=\:9a^2 +6a + 3 \:=\: \underbrace{3(3a^2+2a+1)}_{\text{multiple of 3}}

    (3)\;\;n \:=\:3a+1
    . . . \text{Then: }\;n^2 \;+\; 2n\;+\;3 \;=\;(3a+1)^2 + 2(3a+1)\;+\; 3 \;=\; 9a^2 + 12a \;+\; 6 \;=\;\underbrace{3(3a^2 + 4a+2)}_{\text{multiple of 3}}

    . . Q.E.D.

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