If $y=3 x+6 x^2+10 x^3+\ldots \ldots \ldots \infty$, then $\frac{1}{3} y-\frac{1.4}{3^2 2!} y^2+\frac{1.4 .7}{3^2 3!} y^3-\ldots \ldots \infty$ is equal to

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If \(y=3 x+6 x^2+10 x^3+\ldots \ldots \ldots \infty\), then \(\frac{1}{3} y-\frac{1.4}{3^2 2!} y^2+\frac{1.4 .7}{3^2 3!}…

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