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Math Help - Binomial Theorem

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    Binomial Theorem

    The first four terms in the expansion of (1+px)^{n}, where n>0 are 1+qx+66k^{2}x^{2}+5940x^{3}. Calculate the value of n, of p and of q.

    Thanks!
    Last edited by acc100jt; November 14th 2007 at 06:58 PM.
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    Quote Originally Posted by acc100jt View Post
    The first four terms in the expansion of (1+px)^{n}, where n>0 are 1+qx+66k^{2}x^{2}+5940x^{3}. Calculate the value of n, of p and of q.

    Thanks!
    n has to be 3. That means p^3 = 5940 so p=18.11. Can you do that now?
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  3. #3
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    Never seen that kind of problem before.
    Does it go on? or is that the end at 5940? I'll assume it does end at the highest exponent 3.
    Where does k come from? I'll assume it doesn't matter or it's p, which can be solved after we know what p is.


    For this,
    a= 1
    b= px
    because of the readiness to plug it into pascal's triangle as a and b.

    Pascal's triangle. [(a+b)^3= a^3 + 3a^2*b + 3a*b^2 + b^3

    The problem ends with the highest power being 3 (on x^3 ) which fits the the above part of Pascal's triangle.
    So we now have n.

    So what to the 3rd power equals? 5940
    p^n= 5940

    So now we have p.

    3a^2*b.... is the place where qx exists.

    So q= 3(Px)

    (because 1 which is a to the second power (1^2) is just one and it doesn't change the equation)

    Because p= a in its location in pascal's triangle

    and solve.


    phew!
    Ask if I talked too complicated or missed a mental/verbal/mathematical/grammatical/social/economical step.
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    Take note that n is not 3. The question says "first four terms". n can be any number.

    Anyway, the answer is n=12, p=3, q=36
    Can anyone show me how to get the answer?
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    Quote Originally Posted by acc100jt View Post
    The first four terms in the expansion of (1+px)^{n}, where n>0 are 1+qx+66k^{2}x^{2}+5940x^{3}. Calculate the value of n, of p and of q.
    Thanks!
    Where did those extra terms come from, the q & k?
    Please correct the problem.
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