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Math Help - Pascal's Trianglw

  1. #1
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    Pascal's Trianglw

    Oops, sorry for the typo in the title. :|

    Need help with these please:

    In the expansion of (1-ax)^n, the first three terms are 1-12x-63x^2. Find the values of a and n.

    and

    In the expansion of (x-(1/x^2))^8 find:

    a) The term that has x^-1

    b) The coefficient of the term that has x^2.

    Just to make the second one a little clearer, here's something I whipped up with my MSPaint skills:
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  2. #2
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    Is there a typo in #1. Should it be +63 and not –63?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Is there a typo in #1. Should it be +63 and not Ė63?
    Ah, crud. You're right. 2 Typos in one post.
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  4. #4
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    Quote Originally Posted by StackFryer View Post
    Oops, sorry for the typo in the title. :|

    Need help with these please:

    In the expansion of (1-ax)^n, the first three terms are 1-12x+63x^2. Find the values of a and n.
    (1-ax)^n=1 + n(-ax) + n(n-1)(-ax)^2/2 + ..

    so:

    a.n=12

    and:

    (n(n-1)a^2)/2=63.

    Expanding this last equation:

    (na)^2-(na)a=126,

    or:

    144-12a=126

    so:

    a=3/2

    hence:

    n=8.

    RonL
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  5. #5
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    Quote Originally Posted by StackFryer View Post
    In the expansion of (x-(1/x^2))^8 find:

    a) The term that has x^-1

    b) The coefficient of the term that has x^2.

    Just to make the second one a little clearer, here's something I whipped up with my MSPaint skills:
    rearrange (x-(1/x^2))^8, to:

    (1/x^16) (1-x^3)^8 = (1/x^16)[1+8x^3+28x^6+54x^9+70x^12+54x^15+

    ...................28x^18+8x^21+x^24]

    a) so the coefficient of x^(-1) in this is the coefficient of x^15 in the []'s,
    so is 54.

    b) so the coefficient of x^2 in this is the coefficient of x^18 in the []'s,
    so is 28.

    RonL
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