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Question
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
Options
10
100
150
none of these
Solution
100
\[\lim_{x \to 1} \frac{x + x^2 + x^3 . . . x^n - n}{x - 1} = 5050\]
\[ \Rightarrow \lim_{x \to 1} \frac{\left( x - 1 \right)}{x - 1} + \frac{\left( x^2 - 1 \right)}{x - 1} + \frac{\left( x^3 - 1 \right)}{x - 1} . . . \frac{x^n - 1}{x - 2} = 5050\]
\[ \Rightarrow 1 + 2 + 3 . . . n = 5050 \left[ \because \frac{x^n - a^n}{x - a} = n a^{n - 1} \right]\]
\[ \Rightarrow \frac{n\left( n + 1 \right)}{2} = 5050\]
\[ \Rightarrow n\left( n + 1 \right) = 10100\]
\[ \Rightarrow n\left( n + 1 \right) = 100 \times 101\]
\[\text{ On comparing }: \]
\[n = 100\]
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