If $\mathrm{f}(x)=\frac{x+x^2+x^3+\ldots \ldots \ldots \ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}$, for $x
eq 1$ is continuous at $x=1$, then $\mathrm{f}(1)=$

Question

If $\mathrm{f}(x)=\frac{x+x^2+x^3+\ldots \ldots \ldots \ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}$, for $x 
eq 1$ is continuous at $x=1$, then $\mathrm{f}(1)=$

Aicharya · Accepted Answer

$\begin{aligned} \mathrm{f}(1) & =\lim _{x ightarrow 1} \mathrm{f}(x) \ & =\lim _{x ightarrow 1} \frac{x+x^2+x^3+\ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1} \ & =\lim _{x ightarrow 1} \frac{x-1+x^2-1+x^3-1+\ldots+x^{\mathrm{n}}-1}{x-1}\end{aligned}$ $\begin{aligned} & =\lim _{x ightarrow 1} \frac{x-1}{x-1}+\lim _{x ightarrow 1} \frac{x^2-1}{x-1}+\ldots+\lim _{x ightarrow 1} \frac{x^{\mathrm{n}}-1}{x-1} \ & =1+2(1)^{2-1}+3(1)^{3-1}+\ldots+\mathrm{n}(1)^{\mathrm{n}-1} \ & =1+2+3+\ldots \mathrm{n} \ \mathrm{f}(1) & =\frac{\mathrm{n}(\mathrm{n}+1)}{2}\end{aligned}$

If $\mathrm{f}(x)=\frac{x+x^2+x^3+\ldots \ldots \ldots \ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}$, for $x \neq…

Solution