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Thread: Power Series

  1. #1
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    Power Series

    A dont understand how the step highlighted in red came about. How did the $\displaystyle 3 $ become $\displaystyle 3^2$ and the $\displaystyle 3^2$ become $\displaystyle 3^4$? And, also, how did the $\displaystyle 3^\frac{1}{3}$ before $\displaystyle (1+\frac{y^2}{3})^\frac{1}{3}$ come about when changing the form of $\displaystyle (3+y^2)^\frac{1}{3}$?
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  2. #2
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    Quote Originally Posted by Haris View Post
    A dont understand how the step highlighted in red came about. How did the $\displaystyle 3 $ become $\displaystyle 3^2$ and the $\displaystyle 3^2$ become $\displaystyle 3^4$? And, also, how did the $\displaystyle 3^\frac{1}{3}$ before $\displaystyle (1+\frac{y^2}{3})^\frac{1}{3}$ come about when changing the form of $\displaystyle (3+y^2)^\frac{1}{3}$?
    Hi

    $\displaystyle \left(3+y^2\right)^\frac{1}{3} = \left(3\:\left(1+\frac{y^2}{3}\right)\right)^\frac {1}{3} = 3^\frac{1}{3} \left(1+\frac{y^2}{3}\right)^\frac{1}{3}$

    because $\displaystyle (ab)^n = a^n\:b^n$

    Then

    $\displaystyle \binom{\frac{1}{3}}{1} = \frac{1}{3}$

    Therefore
    $\displaystyle \binom{\frac{1}{3}}{1} \frac{1}{3} \:y^2= \frac{y^2}{9}$

    Then
    $\displaystyle \binom{\frac{1}{3}}{2} = \frac{\frac{1}{3}\:\left(\frac{1}{3}-1\right)}{2} = \frac{\frac{1}{3}\:\left(\frac{-2}{3}\right)}{2} = -\frac{1}{9}$
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