Assertion (A):  $\mathrm{x}+\frac{{\mathrm{x}}^2}{1!}+\frac{{\mathrm{x}}^3}{2!}+\frac{{\mathrm{x}}^4}{3!}+\dots .+\infty =\text{⨍ (x)}$ 

Then the series $\text{⨍(1), ⨍(2), ⨍(3), ⨍(4)….}$ are in G.P.

Reason (R) : In the above series ${\text{⨍ "(x)=e}}^{\mathrm{x}}\text{(x+2)}$

$$\begin{gathered} f\left(\mathrm{x}\right)=\mathrm{x}+\frac{{\mathrm{x}}^2}{\overline{)\underline{1}}}+\frac{{\mathrm{x}}^3}{\overline{)\underline{2}}}+\frac{{\mathrm{x}}^4}{\overline{)\underline{3}}}+\cdots \cdots \\ =\mathrm{x}\left(1+\frac{\mathrm{x}}{\overline{)\underline{1}}}+\frac{{\mathrm{x}}^2}{\overline{)\underline{2}}}+\frac{{\mathrm{x}}^3}{\overline{)\underline{3}}}+\cdots \cdots \right) \\ f\left(\mathrm{x}\right)=\mathrm{x} {\mathrm{e}}^{\mathrm{x}} \\ f\left(1\right)=\mathrm{e}, f\left(2\right)=2{\mathrm{e}}^2, f\left(3\right)=3{\mathrm{e}}^3, f\left(4\right)=4{\mathrm{e}}^4 \\ \therefore \frac{f\left(2\right)}{f\left(1\right)}=2\mathrm{e}, \frac{f\left(3\right)}{f\left(2\right)}=\frac{3\mathrm{e}}{2} \end{gathered}$$

they are not in G.P.

$$\begin{gathered} f^1\left(\mathrm{x}\right)={\mathrm{xe}}^{\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}=(\mathrm{x}+1) {\mathrm{e}}^{\mathrm{x}} \\ {f}^{11}\left(\mathrm{x}\right)=(\mathrm{x}+1) {\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}\left(1\right) \\ =(\mathrm{x}+2) {\mathrm{e}}^{\mathrm{x}} \end{gathered}$$