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k=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128

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Carcass
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k=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128 [#permalink] New post   31 Dec 2021, 07:22
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\(k=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)


Quantity A
Quantity B
\(k\)
\(2\)



The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given
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B
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Re: k=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128 [#permalink] New post   31 Dec 2021, 07:31
Carcass wrote:
\(k=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)


Quantity A
Quantity B
\(k\)
\(2\)



The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given


Given: \(k=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

Rewrite with common denominator: \(k=1+\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128} = 1 + \frac{127}{128}\)

Answer: B
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Re: k=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128 [#permalink] New post   31 Dec 2021, 12:22
The addition starts at 1, and fractions that are decreasing are successively added to it. So, definitely, 2 is greater than the expression in Column A.
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Re: k=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128 [#permalink]
31 Dec 2021, 12:22
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