If y = 3x + 6x^2 + 10x^3 + dots,then the valu | vedclass

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(D) Given the series $y = 3x + 6x^2 + 10x^3 + \dots$Adding $1$ to both sides,we get $1 + y = 1 + 3x + 6x^2 + 10x^3 + \dots$We recognize the right side

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$1 - (1 + y)^{-1/3}$

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(D) Given the series $y = 3x + 6x^2 + 10x^3 + \dots$Adding $1$ to both sides,we get $1 + y = 1 + 3x + 6x^2 + 10x^3 + \dots$We recognize the right side as the binomial expansion of $(1 - x)^{-3}$,which is $1 + (-3)(-x) + \frac{(-3)(-4)}{2!}(-x)^2 + \frac{(-3)(-4)(-5)}{3!}(-x)^3 + \dots = 1 + 3x + 6x^2 + 10x^3 + \dots$Thus,$1 + y = (1 - x)^{-3}$.Taking the power of $-1/3$ on both sides,we get $(1 + y)^{-1/3} = 1 - x$.Rearranging for $x$,we obtain $x = 1 - (1 + y)^{-1/3}$.

If $y = 3x + 6x^2 + 10x^3 + \dots$,then the value of $x$ in terms of $y$ is

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