Question:
The value of \(\lim_{x \to 1} \frac{\sum_{k=1}^{100} x^k - 100}{x - 1}\) is
The value of \(\lim_{x \to 1} \frac{\sum_{k=1}^{100} x^k - 100}{x - 1}\) is
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Use \(\lim_{x \to 1} \frac{x^n - 1}{x - 1} = n\).
Updated On: Apr 23, 2026
- -5050
- 0
- 5050
- None of these
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ \lim_{x \to 1} \frac{x^k - 1}{x - 1} = k \]
Step 2: Calculation / Simplification}
\[ \sum_{k=1}^{100} x^k - 100 = \sum_{k=1}^{100} (x^k - 1) \]
\[ \lim_{x \to 1} \frac{\sum_{k=1}^{100} (x^k - 1)}{x - 1} = \sum_{k=1}^{100} \lim_{x \to 1} \frac{x^k - 1}{x - 1} = \sum_{k=1}^{100} k \]
\[ = \frac{100 \times 101}{2} = 5050 \]
Step 3: Final Answer
\[ 5050 \]
\[ \lim_{x \to 1} \frac{x^k - 1}{x - 1} = k \]
Step 2: Calculation / Simplification}
\[ \sum_{k=1}^{100} x^k - 100 = \sum_{k=1}^{100} (x^k - 1) \]
\[ \lim_{x \to 1} \frac{\sum_{k=1}^{100} (x^k - 1)}{x - 1} = \sum_{k=1}^{100} \lim_{x \to 1} \frac{x^k - 1}{x - 1} = \sum_{k=1}^{100} k \]
\[ = \frac{100 \times 101}{2} = 5050 \]
Step 3: Final Answer
\[ 5050 \]
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