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Math Help - Binomial Theorem for (3n+ 1)^3

  1. #1
    Junior Member
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    Binomial Theorem for (3n+ 1)^3

    I'm a complete beginner at these:

    Could someone please correct me if I'm wrong please (and show me where)....

    Here goes:

    (3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)^2 + 1

    which would give me

    = 3 (9n^3 + 6n^2) +1 is this correct?
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  2. #2
    Junior Member
    Joined
    Dec 2005
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    Quote Originally Posted by Natasha
    I'm a complete beginner at these:

    Could someone please correct me if I'm wrong please (and show me where)....

    Here goes:

    (3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)^2 + 1

    which would give me

    = 3 (9n^3 + 6n^2) +1 is this correct?


     (3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n)1^2 + 1
     (3n+1)^3 = (3n)^3 + 3(3n)^2 + 3(3n) + 1

    I have marked with red ^2 in your post because in theorem its b^2 and because b=1 then b^2 =1, you have put (ab)^2 or a^2.
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  3. #3
    Junior Member
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    Oct 2005
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    thanks :-)
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