mathrm{b}=1+frac{{ }^1 mathrm{C}_0+{ }^1 math | vedclass

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Question

$ \mathrm{f}(\mathrm{x})=1+\frac{(1+\mathrm{x})}{1 !}+\frac{(1+\mathrm{x})^2}{2 !}+\frac{(1+\mathrm{x})^3}{3 !}+\ldots . . $

$ \frac{\mathrm{e}^{(1

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$ \mathrm{f}(\mathrm{x})=1+\frac{(1+\mathrm{x})}{1 !}+\frac{(1+\mathrm{x})^2}{2 !}+\frac{(1+\mathrm{x})^3}{3 !}+\ldots . . $

$ \frac{\mathrm{e}^{(1+\mathrm{x})}}{1+\mathrm{x}}=\frac{1}{1+\mathrm{x}}+1+\frac{(1+\mathrm{x})}{2 !}+\frac{(1+\mathrm{x})^2}{3 !}+\frac{(1+\mathrm{x})^2}{4 !} $

$ 	ext { coef } \mathrm{x}^2 	ext { in RHS : } 1+\frac{{ }^2 \mathrm{C}_2}{3}+\frac{{ }^3 \mathrm{C}_2}{4}+\ldots=\mathrm{a} $

$ 	ext { coeff. } \mathrm{x}^2 	ext { in L.H.S. } $

$ \mathrm{e}\left(1+\mathrm{x}+\frac{\mathrm{x}^2}{2 !}ight) \ldots .\left(1-\mathrm{x}+\frac{\mathrm{x}^2}{2 !} \ldots \ldots .ight) $

$ 	ext { is } \mathrm{e}-\mathrm{e}+\frac{\mathrm{e}}{2 !}=\mathrm{a} $

$ \mathrm{b}=1+\frac{2}{1 !}+\frac{2^2}{2 !}+\frac{2^3}{3 !}+\ldots \ldots=\mathrm{e}^2 $

$ \frac{2 \mathrm{~b}}{\mathrm{a}^2}=8$

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