If f(x) = frac{x+x^2+x^3+ldots+x^{n}-n}{x-1} | vedclass

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(B) Since $f(x)$ is continuous at $x=1$,we have $f(1) = \lim_{x \rightarrow 1} f(x)$.$\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} \frac{x+x^2

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$\frac{n(n+1)}{2}$

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(B) Since $f(x)$ is continuous at $x=1$,we have $f(1) = \lim_{x \rightarrow 1} f(x)$.$\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} \frac{x+x^2+x^3+\ldots+x^{n}-n}{x-1}$We can rewrite the numerator as $(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)$.So,$\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} \left[ \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right]$Using the standard limit $\lim_{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,we get:$\lim_{x \rightarrow 1} f(x) = 1 + 2(1)^{2-1} + 3(1)^{3-1} + \ldots + n(1)^{n-1}$$\lim_{x \rightarrow 1} f(x) = 1 + 2 + 3 + \ldots + n$The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.Therefore,$f(1) = \frac{n(n+1)}{2}$.

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

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