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Math Help - Binomial expansion and differentiation

  1. #1
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    Binomial expansion and differentiation

    1.The first three terms of the expansion of (1+kx)^n are 1+6x+27x^2. Find k and n and hence find the term in x^3.

    2.In the expansion of (1+ax)^n, in ascending powers of x, the coefficient of x is 12 and the coefficient of x^3 is six times the coefficient of x^2. Find a and n.

    3.Find dy/dx in terms of x and y: 2sin2xcos3y=1

    Could anyone help me with the above questions. Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Confuzzled? View Post
    1.The first three terms of the expansion of (1+kx)^n are 1+6x+27x^2. Find k and n and hence find the term in x^3.
    <br />
(1+kx)^n=1+n(kx)+\frac{n(n-1)}{2}(kx)^2+...<br />

    So n\,k=6, and k^2\,(n^2-n)/2=27.

    Taking the second of these, we may rewite it as:

    ((n\,k)^2-(nk)\,k)/2=(36-6k)/2=27, so k=-3, and so n=-2.

    RonL
    Last edited by CaptainBlack; February 6th 2007 at 11:59 AM. Reason: typo correction
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  3. #3
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    Quote Originally Posted by Confuzzled? View Post
    1.The first three terms of the expansion of (1+kx)^n are 1+6x+27x^2. Find k and n and hence find the term in x^3.

    3.Find dy/dx in terms of x and y: 2sin2xcos3y=1
    if:

    2 \sin(2x) \cos(3y)=1,

    then:

    <br />
\frac{d}{dx} [\sin(2x) \cos(3y)]=0<br />

    Then using the product rule:

    <br />
2 \cos(2x) \cos(3y) - 3 \sin(2x) \sin(3y) \frac{dy}{dx}=0<br />

    so rearranging gives:

    \frac{dy}{dx}=\frac{2 \cos(2x) \cos(3y)}{3 \sin(2x) \sin(3y) }=(2/3) \cot(2x) \cot(3y)

    RonL
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  4. #4
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    Hello, Confuzzled?!

    Here's #3 . . .


    3. Find \frac{dy}{dx} in terms of x and y: . 2\sin2x\cos3y\:=\:1

    Differentiate implicitly: . 2\sin2x\cdot\left[\sin3y(-3)\frac{dy}{dx}\right] + 2\cdot\left[ 2\cos2x\right]\cdot\cos3y \:=\:0

    Then: . \frac{dy}{dx}\:=\:\frac{-4\cos2x\cos3y}{-6\sin2x\sin3y} \:=\:\frac{2}{3}\!\cdot\!\frac{\cos2x}{\sin2x}\!\c  dot\!\frac{\cos3y}{\sin3y} \:=\:\frac{2}{3}\cot2x\cot3y<br /> <br />


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  5. #5
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    thank you guys for your help although I'm afraid to say that the answers you got for the first question Captain black are different to the answers that are in the back of the book.
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  6. #6
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    Quote Originally Posted by Confuzzled? View Post
    thank you guys for your help although I'm afraid to say that the answers you got for the first question Captain black are different to the answers that are in the back of the book.
    Type now corrected

    RonL
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  7. #7
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    Hello again, Confuzzled!

    1.The first three terms of the expansion of (1+kx)^n are: . 1+6x+27x^2
    Find k and n and hence find the coefficient of x^3

    Binomial Theorem: . (1 + kx)^n \;=\;\binom{n}{0}1^n + \binom{n}{1}(kx) + \binom{n}{2}(kx)^2 + \binom{n}{3}(kx)^3 + \cdots

    Hence, we have: .[1] \binom{n}{1}k = 6 .and [2] . \binom{n}{2}k^2 = 27

    [1] becomes: . nk \,=\,6 [3]

    [2] becomes: . \frac{n(n-1)}{2}k^2 \,= \,27\quad\Rightarrow\quad n(n-1)k^2 \,=\,54 [4]

    Divide [4] by [3]: . \frac{n(n-1)k^2}{nk} \:=\:\frac{54}{6}\quad\Rightarrow\quad(n-1)k \:=\:9

    Divide by [3]: . \frac{(n-1)k}{nk} \:=\:\frac{9}{6}\quad\Rightarrow\quad\frac{n-1}{n} \,=\,\frac{3}{2}

    Solve for n\!:\;\;2n - 2 \:=\:3n\quad\Rightarrow\quad\boxed{n = -2}

    Substitute into [3]: . -2k \,=\,6\quad\Rightarrow\quad\boxed{k \,=\,-3}

    Then: . \binom{n}{3}k^3 \:=\:\frac{(-2)(-3)(-4)}{6}(-3)^3\:=\:\boxed{108} . . . coefficient of x^3

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Check

    (1 - 3x)^{-2} \;=\;\frac{1}{1 - 6x + 9x^2} \;=\;1 + 6x + 27x^2 + 108x^3 + \cdots

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