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Math Help - Using Pascal's Triangle to Solve a sqrt Binomial

  1. #1
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    Using Pascal's Triangle to Solve a sqrt Binomial

    I need an answer to these expressions using the Pascal method... I was doing very well on my work until I ran into these two questions:

    <br />
(\sqrt{a}+\sqrt{b})^6<br />

    and

    <br />
(x + 1/x)^4<br />

    If anyone could help me on these, it'd be greatly appreciated.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by aMUSEment View Post
    I need an answer to these expressions using the Pascal method... I was doing very well on my work until I ran into these two questions:

    <br />
(\sqrt{a}+\sqrt{b})^6<br />

    and

    <br />
(x + 1/x)^4<br />

    If anyone could help me on these, it'd be greatly appreciated.
    Pascal's triangle only works for the coefficients of each term and I've never heard of Pascal's method. The easiest way would be the binomial theorem

    (a+b)^n = \sum_{k=0}^n {n \choose k}a^{n-k}b^k

    Where {n \choose k} = \frac{n!}{(n-k)!k!}


    Taking your first example (n=6) I will take the first two terms

    (a^{\frac{1}{2}} + b^{\frac{1}{2}})^6 = {6 \choose 0} (a^{\frac{1}{2}})^3 (b^{\frac{1}{2}})^0 + {6 \choose 1} (a^{\frac{1}{2}})^{6-1} (b^{\frac{1}{2}})^1

    Simplifying that we get a^3 + 6a^{\frac{5}{2}}b^{\frac{1}{2}}


    Have a go at the rest of it and post if you don't understand
    Last edited by e^(i*pi); May 22nd 2010 at 01:59 PM. Reason: latex
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