JEE Main 2020 (Online) 2nd September Morning ShiftQuestion 1
If $\lim_{x\to1}\frac{x+x^2+x^3+\ldots+x^n-n}{x-1}=820$, ($n\in\mathbf{N}$) then the value of $n$ is equal to ________.
Limits & Continuity
Answer$40$
Solution
Concept. Write $-n$ as $-1$ taken $n$ times and pair each with a power. Formulas used. (1) $\lim_{x\to1}\frac{x^i-1}{x-1}=i$(2) $1+2+\dots+n=\frac{n(n+1)}{2}$Step 1 - Split. $$\frac{x+x^2+\dots+x^n-n}{x-1}=\sum_{i=1}^{n}\frac{x^i-1}{x-1}$$ Step 2 - Limit (1), (2). $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$. Step 3 - Solve. $\frac{n(n+1)}{2}=820\Rightarrow n(n+1)=1640=40\cdot41\Rightarrow n=40$. Answer: 40