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Math Help - 1 / square root X binomial expansion

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    1 / square root X binomial expansion

    I have this problem and it involves 1/square root X

    I have this nagging feeling theres somthing AI have to do with it but I can't get it

    The problem is:

    Write the 1st 3 terms in each expansion in simplified form:
    (x - 1/square root X) to the power of 5
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by geomeTRY View Post
    I have this problem and it involves 1/square root X

    I have this nagging feeling theres somthing AI have to do with it but I can't get it

    The problem is:

    Write the 1st 3 terms in each expansion in simplified form:
    (x - 1/square root X) to the power of 5
    this is Binomial expansion. there are two main ways to do it. we can use the Pascal's triangle method or the Binomial theorem method. which have you learnt?
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    Hello, geomeTRY!

    Write the 1st 3 terms: . \left(x^2 -\frac{1}{\sqrt{x}}\right)^5
    I must assume you know the Binomial Theorem . . .

    We have: . {5\choose5}(x^2)^5 \;-\; {5\choose4}(x^2)^4\left(\frac{1}{\sqrt{x}}\right) \;+\; {5\choose3}(x^2)^3\left(\frac{1}{\sqrt{x}}\right)^  2 ;+\; \cdots

    . . = \;1\cdot x^{10} \:-\: 5\cdot x^8 \cdot \frac{1}{x^{\frac{1}{2}}} \:+\: 10 \cdot x^6\cdot\frac{1}{x}

    . . = \;x^{10} \,-\, 5x^{\frac{15}{2}} \,+\, 10x^5

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    Quote Originally Posted by geomeTRY View Post
    I have this problem and it involves 1/square root X

    I have this nagging feeling theres somthing AI have to do with it but I can't get it

    The problem is:

    Write the 1st 3 terms in each expansion in simplified form:
    (x - 1/square root X) to the power of 5
    First do (a+b)^5

    a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5

    Now let a=x^2 \mbox{ and } b = -\frac{1}{\sqrt{x}} = -x^{-1/2}

    We get,

    (x^2)^5 + 5(x^2)^4(-x^{-1/2}) +10 (x^2)^3 (-x^{-1/2})^2 + 10 (x^2)^2 (-x^{-1/2})^3 +5(x^2)(-x^{-1/2})^4 + (-x^{-1/2})^5


    Now I let you simplify that.
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  5. #5
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    i learned it that you expand it to

    tn = C(n, r) (a to the power of n-r) (b to the power of r)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by geomeTRY View Post
    i learned it that you expand it to

    tn = C(n, r) (a to the power of n-r) (b to the power of r)
    Yes, that's the binomial theorem:

    (a + b)^n = \sum_{r=0}^{n} {n \choose r} a^{n-r}b^r
    Last edited by Jhevon; June 6th 2007 at 08:44 PM.
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