Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+{\mathrm{x}}^2+\dots \dots +{\mathrm{x}}^{\mathrm{n}}-\mathrm{n}}{\mathrm{x}-1}, \mathrm{x}\ne 1$ the value of f(1) so that f is continuous at x = 1 is

$$\begin{gathered} \begin{array}{l}\underset{\mathrm{x}\to 1}{\mathrm{Lt}}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(1\right)\mathrm{f}\left(1\right)=\underset{\mathrm{x}\to 1}{\mathrm{Lt}}\frac{\mathrm{x}+{\mathrm{x}}^2+\dots \dots +{\mathrm{x}}^{\mathrm{n}}-\mathrm{n}}{\mathrm{x}-1},\left(\frac{0}{0}\text{ from }\right)\\ =\underset{\mathrm{x}\to 1}{\mathrm{Lt}}\frac{1+2\mathrm{x}+3{\mathrm{x}}^2\dots \dots +{\mathrm{nx}}^{\mathrm{n}+1}-0}{1-0}=1+2+3+\dots .+\mathrm{n} \\ \mathrm{f}\left(1\right)=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\end{array} \end{gathered}$$