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Thread: [SOLVED] Binomial expansion problem

  1. #1
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    Question [SOLVED] Binomial expansion problem

    I'm having problem with my revision sheets - Any help would be much appreciated...

    (i) Write down the general term in the power series expansion of
    ((x^2) - (2/x))^18
    WHAT IS THE GENERAL TERM AND HOW WOULD I GO ABOUT THIS?

    Hence find the term which is independant of x
    I DONT KNOW WHAT THIS MEANS

    (ii)Find the first four terms in the expansion of (1+x)^16 in ascending powers of x.

    hence expand (1 + z + z^2) in ascending powers of z up to and including the term in z^3

    DO YOU FACTORISE THIS AND RAISE BOTH PARENTHESIS TO THE POWER 16??
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  2. #2
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    Hello dojo
    Quote Originally Posted by dojo View Post
    I'm having problem with my revision sheets - Any help would be much appreciated...

    (i) Write down the general term in the power series expansion of
    ((x^2) - (2/x))^18
    WHAT IS THE GENERAL TERM AND HOW WOULD I GO ABOUT THIS?

    Hence find the term which is independant of x
    I DONT KNOW WHAT THIS MEANS

    (ii)Find the first four terms in the expansion of (1+x)^16 in ascending powers of x.

    hence expand (1 + z + z^2) in ascending powers of z up to and including the term in z^3

    DO YOU FACTORISE THIS AND RAISE BOTH PARENTHESIS TO THE POWER 16??
    (i) Take out a factor $\displaystyle x^2$, whence it becomes $\displaystyle \big(x^2\big)^{18}$:
    $\displaystyle \left(x^2-\frac2x\right)^{18} = x^{36}\left(1-\frac{2}{x^3}\right)^{18}$
    Now let $\displaystyle y = -\frac{2}{x^3}$, and expand $\displaystyle (1+y)^{18}$, the general term of which is $\displaystyle \binom{18}{r}y^r$, to get the general term:
    $\displaystyle x^{36}\binom{18}{r}\left(-\frac{2}{x^3}\right)^r=(-2)^r\binom{18}{r}x^{36-3r}$
    The term that's independent of $\displaystyle x$ is the one where the power of $\displaystyle x$ is zero. So put $\displaystyle r = 12$ ...

    (ii)
    $\displaystyle (1+x)^{16} = 1 + 16x+ \binom{16}{2}x^2 + \binom{16}{3}x^3+ ...$
    Now put $\displaystyle x = z+z^2$:
    $\displaystyle (1+z+z^2)^{16} = 1 + 16(z+z^2)+ \binom{16}{2}(z+z^2)^2 + \binom{16}{3}(z+z^2)^3+ ...$
    = ...
    Expand each $\displaystyle (z+z^2)$ term up to the term in $\displaystyle z^3$; collect like terms, and you're done.

    Grandad
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  3. #3
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    Hello, dojo!

    (i) Find the general term in the expansion of: .$\displaystyle \left(x^2 - \frac{2}{x}\right)^{18}$
    Assuming that you know the Binomial Theorem, this is easy:

    . . $\displaystyle {18\choose n}\left(x^2\right)^n\left(\text{-}\frac{2}{x}\right)^{18-n}\;\;\text{ for }n \,=\,18,17,16\,\hdots 0 $



    Hence, find the term which is independant of $\displaystyle x$.
    $\displaystyle {\color{blue}\text{It means "Find the term which has no }x."}$
    Somewhere in the middle, we have: .$\displaystyle {18\choose6}\left(x^2\right)^6\left(\text{-}\frac{2}{x}\right)^{12} $

    . . $\displaystyle =\; \frac{18!}{6!\,12!}\left(x^{12}\right)\left(\frac{ (\text{-}2)^{12}}{x^{12}}\right) \;=\;(18,\!564)\left(x^{12}\right)\left(\frac{4,\! 096}{x^{12}}\right) \;=\; 76,\!038,\!144 $

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  4. #4
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    Thumbs up

    Thank you both for the insight very much appreciated!
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