Question

If $\underset{x\to 1}{lim}\frac{x+x^2+x^3+......+x^n-n}{x-1}=820$, $n\in N$, then the value of $n$ is equal to:

Answer 1

Step 1: Apply L' Hospital's Rule.

By applying limit, we get

$\frac{\left(1+1^2+1^3+......nterms\right)-n}{1-1}=\frac{0}{0}$, so we will apply L' Hospital's Rule.

According to this rule $\underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$.

So,

$\begin{array}{rcl}\underset{x\to 1}{lim}\frac{x+x^2+x^3+......+x^n-n}{x-1}& =& \underset{x\to 1}{lim}\frac{1+2x+3x+....+n{x}^{n-1}}{1}\\ & =& 1+2+3+......+n\end{array}$

Step 2: Find the value of $n$.

We have $\underset{x\to 1}{lim}\frac{x+x^2+x^3+......+x^n-n}{x-1}=820$

$\begin{array}{rcl}& \Rightarrow & 1+2+3+...+n=820\\ \Rightarrow \frac{n\left(n+1\right)}{2}& =& 820\left(\mathrm{by}\mathrm{sum}\mathrm{of}\mathrm{first}n\mathrm{natural}\mathrm{numbers}\right)\\ \Rightarrow n^2+n-1640& =& 0\end{array}$

By applying the formula of root of a quadratic equation, we get

$\begin{array}{rcl}n& =& \frac{-1\pm \sqrt{1+6560}}{2}\\ & =& \frac{-1\pm 81}{2}\\ & =& \frac{80}{2}\left(\mathrm{ignoring}\mathrm{the}-ve\mathrm{value}\right)\\ & =& 40\end{array}$

Therefore, $n=40$.