Question:
Let \(a_{n} = 2^{n - 1}, n = 1, 2, 3, \ldots\) . Then the value of the sum \(\sum_{n = 1}^{20} a_{n}\) is equal to
Let \(a_{n} = 2^{n - 1}, n = 1, 2, 3, \ldots\) . Then the value of the sum \(\sum_{n = 1}^{20} a_{n}\) is equal to
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\(\sum_{k=0}^{n-1} 2^k = 2^n - 1\).
Updated On: Apr 25, 2026
- $\dfrac{2^{20} - 1}{2^5}$
- $\dfrac{2^{21} - 1}{\sqrt{2}}$
- $2^{20} - 1$
- $\dfrac{2^{21} - 1}{2^{10}}$
- $\dfrac{2^{20} - 1}{\sqrt{2}(2^{19} - 1)}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Geometric series: \(a_n = 2^{n-1}\), first term 1, ratio 2, n=20 terms.
Step 2: Detailed Explanation:
Sum = \(\frac{1(2^{20} - 1)}{2 - 1} = 2^{20} - 1\).
Step 3: Final Answer:
Option (c).
Geometric series: \(a_n = 2^{n-1}\), first term 1, ratio 2, n=20 terms.
Step 2: Detailed Explanation:
Sum = \(\frac{1(2^{20} - 1)}{2 - 1} = 2^{20} - 1\).
Step 3: Final Answer:
Option (c).
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