$y=3 x+6 x^{2}+10 x^{3}+\ldots \infty$ হলে দেখাও যে,$x=\frac{1}{3} y-\frac{1}{3^{2}} \cdot \frac{4}{2 !} y^{2}+\frac{1.4 .7}{3^{3} \cdot 3 !} y^{3}-\ldots \ldots \ldots$

$\begin{array}{l}\text { } y=3 x+6 x^{2}+10 x^{3}+\ldots \\ \Rightarrow 1+y=1+3 x+6 x^{2}+10 x^{3}+\ldots=(1-x)^{-3} \\ \Rightarrow(1+y)^{-\frac{1}{3}}=(1-x)^{-3 \times\left(-\frac{1}{3}\right)} \\ \Rightarrow 1-x=(1-y) \frac{1}{3} \\ \Rightarrow 1-x=1-\frac{1}{3} y+\frac{\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right) y^{2}\left(-\frac{1}{3}\right)\left(-\frac{1}{3}-1\right)\left(-\frac{1}{3}-2\right)}{2 !} y^{3}+\ldots \\ \therefore x=\frac{1}{3} y-\frac{1}{3^{2}} \cdot \frac{4}{2 !} y^{2}+\frac{1.4 .7}{3^{3} \cdot 3 !} y^{3}-\ldots(\text { showed }) \\\end{array}$