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Question Number 202765 by Mastermind last updated on 02/Jan/24

Answered by aleks041103 last updated on 02/Jan/24

$$\frac{x+x^2+...+x^n-n}{x-1}=$$$$=\underset{k=1}{\sum ^n}\frac{x^k-1}{x-1}=$$$$=\underset{k=1}{\sum ^n}\underset{s=0}{\sum ^{k-1}}{x}^s$$$$\Rightarrow \underset{x\to 1}{lim}\frac{x+x^2+...+x^n-n}{x-1}=\underset{x\to 1}{lim}\underset{k=1}{\sum ^n}\underset{s=0}{\sum ^{k-1}}{x}^s=$$$$=\underset{k=1}{\sum ^n}\underset{s=0}{\sum ^{k-1}}\left(1\right)=\underset{k=1}{\sum ^n}k=\frac{n(n+1)}{2}$$$$\Rightarrow \frac{n(n+1)}{2}=820$$$$\Rightarrow n^2+n-1640=0$$$$n_{1,2}=\frac{-1\pm \sqrt{1+4.1640}}{2}\in \mathbb{N}$$$$\Rightarrow n=\frac{\sqrt{6561}-1}{2}=40$$$$\Rightarrow n=40$$

Answered by manxsol last updated on 02/Jan/24

$$\frac{0}{0}\Rightarrow L^{\prime }Hospital$$$$\underset{x\to 1}{lim}\frac{1+2x+3x^2+.....{nx}^{n-1}}{1}$$$$=\frac{n(n+1)}{2}=820$$$$n(n+1)=41\times 20\times 2=40\times 41$$$$n=40$$

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