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Math Help - Mathematical Induction Question

  1. #1
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    Mathematical Induction Question

    Hi guys, I've been reading the forum for the past 2 semesters of school now and came across a question I couldn't find in the search.

    I am to prove 1/1*2*3 +1/2*3*4 + 1/3*4*5 +...+ 1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2)

    I am at solving the induction part, but am unable to do finish, I keep getting stuck on what seems to be moderate algebra and I don't know what to do next.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by tiddlywinks View Post
    Hi guys, I've been reading the forum for the past 2 semesters of school now and came across a question I couldn't find in the search.

    I am to prove 1/1*2*3 +1/2*3*4 + 1/3*4*5 +...+ 1/n(n+1)(n+2) = n(n+3)/4(n+1)(n+2)

    I am at solving the induction part, but am unable to do finish, I keep getting stuck on what seems to be moderate algebra and I don't know what to do next.

    Thanks in advance!
    I assume you've confirmed the expression is true for n = 1.

    assume true for n ... show true for n+1

    1/1*2*3 +1/2*3*4 + 1/3*4*5 +...+ 1/n(n+1)(n+2) + 1/(n+1)(n+2)(n+3) =

    n(n+3)/[4(n+1)(n+2)] + 1/[(n+1)(n+2)(n+3)] =

    n(n+3)^2/[4(n+1)(n+2)(n+3)] + 4/[4(n+1)(n+2)(n+3)] =

    [n(n+3)^2 + 4]/[4(n+1)(n+2)(n+3)] =

    [n^3 + 6n^2 + 9n + 4]/[4(n+1)(n+2)(n+3)] =

    [(n+1)^2 * (n+4)]/[4(n+1)(n+2)(n+3)] =

    (n+1)(n+4)/[4(n+2)(n+3)] =

    (n+1)[(n+1)+3]/[4[(n+1)+1][(n+1)+2]]
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  3. #3
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    Thanks skeeter. I kept screwing up the n^3 part. Guess it was easy algebra and not moderate!
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