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Thread: Binomial Expansions

  1. #1
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    Binomial Expansions

    Q1)Find, in its simplest form, the coefficient of $\displaystyle x^r$ in the expansion, in ascending powers of $\displaystyle x $ , of $\displaystyle \frac{1}{x-3} $

    Q2) Expand $\displaystyle (1+y)^14 $ as a series of ascending powers of $\displaystyle y $ up to and including the term in $\displaystyle y^3$, Simplify the coefficients.

    I got the above part right and derived at the answer of $\displaystyle 1 + 14y + 91y^2 + 364y^3$

    However, its the following part that i dont get

    In The expansion of $\displaystyle (1 + x + kx^2)^14 $, where $\displaystyle k $ is a constant, the coefficient of $\displaystyle x^3$ is zero. By writing $\displaystyle x + kx^2$ as $\displaystyle y $, or otherwise, find the value of k.

    Any help is appreciated ^^
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  2. #2
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    Quote Originally Posted by teddybear67 View Post
    Q1)Find, in its simplest form, the coefficient of $\displaystyle x^r$ in the expansion, in ascending powers of $\displaystyle x $ , of $\displaystyle \frac{1}{x-3} $

    Q2) Expand $\displaystyle (1+y)^14 $ as a series of ascending powers of $\displaystyle y $ up to and including the term in $\displaystyle y^3$, Simplify the coefficients.

    I got the above part right and derived at the answer of $\displaystyle 1 + 14y + 91y^2 + 364y^3$

    However, its the following part that i dont get

    In The expansion of $\displaystyle (1 + x + kx^2)^14 $, where $\displaystyle k $ is a constant, the coefficient of $\displaystyle x^3$ is zero. By writing $\displaystyle x + kx^2$ as $\displaystyle y $, or otherwise, find the value of k.

    Any help is appreciated ^^
    hi

    begin with some substitution as hinted by the question

    1+14(x+kx^2)+91(x+kx^2)^2+364(x+kx^2)^3+91(x^2+2kx ^3+k^2x^4)+364(x^3+3kx^4+3x(kx^2)^2+(kx^2)^3)

    Given that the coefficient of $\displaystyle x^3$ is 0 , so we need to know the coefficient of x^3 only . To expand the whole thing would be tedious , so we expand those parts where $\displaystyle x^3$ is .

    $\displaystyle ..... 91x^2+182kx^3 + ... +364x^3+....$

    coefficient of x^3 : $\displaystyle 182k+364\Rightarrow k=-2 $
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  3. #3
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    Thanks!
    But do ya know how to do the top one too? :/
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  4. #4
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    $\displaystyle \frac{1}{x- 3}= (x- 3)^{-1}$

    Use the "generalized binomial theorem"
    $\displaystyle (a+ b)^r= \sum_{i=0}^\infty\begin{pmatrix}r \\ i\end{pmatrix}a^ib^{r-i}$
    Where, for r not a positive integer,
    $\displaystyle \begin{pmatrix}r \\ i\end{pmatrix}$$\displaystyle = \frac{r(r-1)\cdot\cdot\cdot(r- i+1)}{i!}$
    With r not a positive integer, r- i+ 1 is never equal to 0 so this is an infinite series.
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