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Math Help - [SOLVED] Binomial Theorem help...

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    Member looi76's Avatar
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    [SOLVED] Binomial Theorem help...

    Question:
    Given that the expansion of (1+ax)^n begins 1+36x+576x^2, find the values of a and n.
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    Quote Originally Posted by looi76 View Post
    Question:
    Given that the expansion of (1+ax)^n begins 1+36x+576x^2, find the values of a and n.
    36x = nax \implies 36 = na.
    575x^2 = \frac{n(n-1)}{2} a^2 x^2 \implies 575 = \frac{n(n-1)}{2} a^2
    Now solve these equations.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    Given that the expansion of (1+ax)^n begins 1+36x+576x^2, find the values of a and n.
    Well
    (1 + ax)^n = \sum_{k = 0}^n {n \choose k}(ax)^k = 1 + n \cdot ax + \frac{n(n - 1)}{2} \cdot a^2x^2 + ~...

    Thus
    na = 36
    and
    \frac{n(n - 1)}{2} \cdot a^2 = 576

    So
    a = \frac{36}{n}

    giving:
    \frac{n(n - 1)}{2} \cdot \left ( \frac{36}{n} \right ) ^2 = 576

    \frac{n - 1}{2n} \cdot 1296 = 576

    Giving me n = 9 and a = 4.

    -Dan
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