Mathematics · 2020
JEE Main · 2 September 2020, Shift 1 · Q72
If lim_x → 1 (x+x^2+x^3+…+x^n - n)/(x-1)=820,( n ∈ N ) then the value of n is equal to \ \ \ \.
If
$$\lim _{x \rightarrow 1} \frac{x+x^2+x^3+\ldots+x^{\mathrm{n}}-\mathrm{n}}{x-1}=820,(\mathrm{n} \in \mathbf{N})
$$
then the value of n is equal to \_\_\_\_.
Official answer
40
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.