1 + frac{1}{3}x + frac{1 cdot 4}{3 cdot 6}x^2 | vedclass

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Question

(D) The general binomial expansion for any index $n$ is given by $(1 + y)^n = 1 + ny + \frac{n(n - 1)}{2!}y^2 + \frac{n(n - 1)(n - 2)}{3!}y^3 + \dots$

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$(1 - x)^{-1/3}$

Vedclass · Answer

(D) The general binomial expansion for any index $n$ is given by $(1 + y)^n = 1 + ny + \frac{n(n - 1)}{2!}y^2 + \frac{n(n - 1)(n - 2)}{3!}y^3 + \dots$Comparing the given series $1 + \frac{1}{3}x + \frac{1 \cdot 4}{3 \cdot 6}x^2 + \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9}x^3 + \dots$ with the expansion:$ny = \frac{1}{3}x$$\frac{n(n - 1)}{2}y^2 = \frac{1 \cdot 4}{3 \cdot 6}x^2 = \frac{4}{18}x^2 = \frac{2}{9}x^2$From $ny = \frac{1}{3}x$,we have $y = \frac{x}{3n}$. Substituting this into the second term:$\frac{n(n - 1)}{2} \cdot \frac{x^2}{9n^2} = \frac{2}{9}x^2$$\frac{n - 1}{18n} = \frac{2}{9} \implies n - 1 = 4n \implies 3n = -1 \implies n = -\frac{1}{3}$Substituting $n = -\frac{1}{3}$ into $ny = \frac{1}{3}x$ gives $(-\frac{1}{3})y = \frac{1}{3}x$,so $y = -x$.Thus,the series is $(1 - x)^{-1/3}$.

$1 + \frac{1}{3}x + \frac{1 \cdot 4}{3 \cdot 6}x^2 + \frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9}x^3 + \dots$ is equal to

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