Discussion
The sum of the geometric series \(\sqrt{3} + \sqrt{12} + \sqrt{48} +\) ...up to 10 terms is
Answer: A
The sum is \(1023\sqrt{3}\).
The terms are \(\sqrt{3}, 2\sqrt{3}, 4\sqrt{3}, \ldots\), since \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{48} = 4\sqrt{3}\). So the first term is \(\sqrt{3}\) and the common ratio is 2:
\[ \begin{aligned} S_{10} &= \sqrt{3}\,\frac{2^{10} - 1}{2 - 1} \\ &= 1023\sqrt{3} \end{aligned} \]
\(1024\sqrt{3}\) forgets the -1 in \(r^{n} - 1\), and \(511\sqrt{3}\) and \(512\sqrt{3}\) use 9 terms instead of 10.
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