Discussion
Let \(a_{n} = 2^{n-1}, n = 1, 2, 3....\) Then the value of the sum \(\sum_{n=1}^{20} a_{n}\) is equal to
Answer: C
The sum is \(2^{20} - 1\).
The terms \(1, 2, 4, \ldots, 2^{19}\) form a G.P. with first term 1 and ratio 2, so
\[ \begin{aligned} \sum_{n=1}^{20} 2^{n-1} &= \frac{2^{20} - 1}{2 - 1} \\ &= 2^{20} - 1 \end{aligned} \]
The same follows from binary: twenty 1s in a row is \(2^{20} - 1\). The other options divide by powers of 2 or \(\sqrt{2}\) that do not appear in the sum, and none of them equals \(2^{20} - 1\).
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