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pre-calculus

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If

(1)(2) + (2)(3) + ... n(n+1) =
n(n+1)(n+2)/3

complete the inductive step of its proof:

1(2) + 2(3) + ... k(k+1) + (k+1)(k+2)

I have no idea what to do.

  • pre-calculus -

    step1
    test for n=1
    LS = (1)(2) = 2
    RS = (1)(2)(3)/3 = 2, check!

    step2
    assume it is true for n=k, that is ...
    (1)(2) + (2)(3) + ... k(k+1) =
    k(k+1)(k+2)/3

    step3
    show that it must be true for n= k+1
    that is ...
    show
    (1)(2) + (2)(3) + ... k(k+1) + (k+1)(k+2) =
    (k+1)(k+2)(k+3)/3

    LS = (1)(2) + (2)(3) + ... k(k+1) + (k+1)(k+2)
    = [ (1)(2) + (2)(3) + ... k(k+1) ] + (k+1)(k+2)
    = k(k+1)(k+2)/3 + (k+1)(k+2)
    = k(k+1)(k+2)/3 + 3(k+1)(k+2)/3
    = [ k(k+1)(k+2) + 3(k+1)(k+2) ]/3
    = (k+1)(k+2) [k + 3]/3 , took out common factor
    = (k+1)(k+2)(k+3)/3
    = RS

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