If $\mathrm{f}\left(\mathrm{x}\right)=(1-\mathrm{x}{)}^{-1},|\mathrm{x}|<1$then the value of $\frac{{\mathrm{f}}^{\mathrm{ii}}\left(0\right)}{\mathrm{f}\left(0\right)}+\frac{{\mathrm{f}}^{\mathrm{iii}}\left(0\right)}{{\mathrm{f}}^{\mathrm{i}}\left(0\right)}+\frac{{\mathrm{f}}^{\mathrm{iv}}\left(0\right)}{{\mathrm{f}}^{\mathrm{ii}}\left(0\right)}+\dots .+\frac{{\mathrm{f}}^{\mathrm{n}+1}\left(0\right)}{{\mathrm{f}}^{\mathrm{n}-1}\left(0\right)}$ is 

Given $\mathrm{f}\left(\mathrm{x}\right)=(1-\mathrm{x})-1,\left|\mathrm{x}\right|<1$ We have on successively differentiating with respect to x

$\begin{array}{ll}\mathrm{f}\left(\mathrm{x}\right)=(1-\mathrm{x}{)}^{-1};\mathrm{f}(0)=1{\mathrm{f}}^{\mathrm{i}}(\mathrm{x})=(1-\mathrm{x}{)}^{-2};{\mathrm{f}}^{\mathrm{i}}\left(0\right)=1{\mathrm{f}}^{\mathrm{ii}}\left(\mathrm{x}\right)=1.2.(1-\mathrm{x}{)}^{-3};{\mathrm{f}}^{\mathrm{ii}}(0)=2!& \\ {\mathrm{f}}^{\mathrm{iii}}\left(\mathrm{x}\right)=1.2.3.(1-\mathrm{x}{)}^{-4};{\mathrm{f}}^{\mathrm{iii}}(0)=3!{\mathrm{f}}^{\mathrm{k}}(\mathrm{x})=\mathrm{k}!(1-\mathrm{x}{)}^{-\mathrm{k}-1};{\mathrm{f}}^{\mathrm{k}}\left(0\right)=\mathrm{k}!& \\ \therefore \frac{{\mathrm{f}}^{\mathrm{ii}}\left(0\right)}{\mathrm{f}\left(0\right)}+\frac{{\mathrm{f}}^{\mathrm{iii}}\left(0\right)}{{\mathrm{f}}^{\mathrm{i}}\left(0\right)}+\frac{{\mathrm{f}}^{\mathrm{iv}}\left(0\right)}{{\mathrm{f}}^{\mathrm{ii}}\left(0\right)}+\dots +\frac{{\mathrm{f}}^{\mathrm{n}+1}\left(0\right)}{{\mathrm{f}}^{\mathrm{n}-1}\left(0\right)}& \mathrm{ }\\ =1.2+2.3+\dots +\mathrm{n}(\mathrm{n}+1)& \\ =\sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}(\mathrm{k}+1)=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}& \end{array}$