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Math Help - need help with induction prove..

  1. #1
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    Post need help with induction prove..

    How do you prove this?
    Use induction to prove that 3 divides n^3 + 3n^2 + 2n for every nonnegative integer n.
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  2. #2
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    Hello, st4rl1ight!

    Use induction to prove that 3 divides n^3 + 3n^2 + 2n for every n > 0
    S(n)\!:\;\;n^3 + 3n^2 + 2n is a multiple of 3.


    Verify S(1)\!:\;\;1^3 + 3\!\cdot\!1^2 + 2\!\cdot\!1 \:=\:6\:=\:3(2) . . . true!


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


    Add (3k^2 + 3k + 1)+(6k+3) + 2 to both sides:

    . . k^3 \;{\color{blue}+ \;3k^2 + 3k + 1} + \;3k^2 \;{\color{blue}+ \;6k + 3} \;+\; 2k \;{\color{blue}+\; 2} \;=\;3a \;+ \;{\color{blue}(3k^2 + 3k + 1) + (6k+3) + 2}


    \text{And we have: }\;(k^3 + 3k^2 + 3k + 1) \;+\;3(k^2 + 2k +1) \;+ \;2(k+1) \;=\;3a \;+ \;3k^2 \;+ \;9k \;+ \;6


    . . (k+1)^3 + 3(k+1)^2 + 2(k+1) \;=\; \underbrace{3\left(a + k^2 + 3k + 2\right)}_{\text{a multiple of 3}}

    We have proved S(k+1) . . . The inductive proof is compete.

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