Question

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

• A. $n$
  
• B. $\frac{n+1}{2}$
• C. $\frac{n(n+1)}{2}$
• D. $\frac{n(n-1)}{2}$

Answer 1

The correct option is C $\frac{n(n+1)}{2}$

$\underset{x\to 1}{lim}f\left(x\right)$$=\underset{x\to 1}{lim}\frac{x+x^2+.....+x^n-n}{x-1}$

$=\underset{x\to 1}{lim}\frac{x-1+x^2-1+.....+x^n-1}{x-1}$

$=\underset{x\to 1}{lim}\frac{x-1}{x-1}+\underset{x\to 1}{lim}\frac{x^2-1}{x-1}+.............+\underset{x\to 1}{lim}\frac{x^n-1}{x-1}$

$=1+2(1{)}^{2-1}+3(1{)}^{3-1}..........n(1{)}^{n-1}$

$=1+2+3+.......n$

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

$f$ is continuous at $x=1$ if

$\underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$

So, $f\left(1\right)=\frac{n(n+1)}{2}$