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Which of the following is the sum of an infinite geometric sequence whose terms come from the set $(1, {1\over 2},{1\over 4},\ldots{1\over 2^n}\ldots) $?

Given options are $a){1\over 5} \ \ b){1\over 7} \ \ c){1\over 9} \ \ d){1\over 11}$

But I got the sum of $(1, {1\over 2},{1\over 4},\ldots {1\over 2^n} \ldots) = {1\over 1-{1\over 2}}=2$. Please help where I've gone wrong. Thanks in advance.

Umberto P.
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Chris
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    The question is asking which of those numbers can be obtained by summing only some of the terms of the sequence. – Umberto P. May 07 '21 at 14:45
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    For exsmple: $1/2 + 1/2^3 + 1/2^5 + \cdots = (1/2)/(1 - 1/4) = 2/3$, so $2/3$ would be a correct answer if it were one of the choices. – Michael Lugo May 07 '21 at 14:49
  • @MichaelLugo that is a very good comment which would make a nice answer to the question. – Umberto P. May 07 '21 at 14:56
  • So first there's the misunderstanding of what the question is asking which was sorted out in the above comments. As for moving forward, I'd suggest "parametrizing" the set of all geometric sequences formed from subsequences of that one, meaning representing them as a function of some small collection of numbers. There is a convenient way to do it using two integers; do you see what two integers they should be? Once you have it parametrized it should be pretty easy to see what sorts of numbers can be represented in this way. – Ian May 07 '21 at 15:06

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Note that the key phrase is geometric sequence. If arbritrary subsequences were allowed, then any real number in the range $[0,2]$ would work. The question is essentially asking which of the above has binary representation which is a geometric series. Geometric sequences of this form must have common ratio $r=\frac{1}{2^n}$ and first term $a=\frac{1}{2^m}$ for some $n>0,m\geq0$, hence sum which can be represented as $$\frac{a}{1-r} = \frac{\frac{1}{2^m}}{1-\frac{1}{2^n}} = \frac{1}{2^m-2^{m-n}}\quad n>0,m\geq0$$ If $m\geq n$, we see that this representation is in lowest terms. If $m<n$, then $$ \frac{1}{2^m-2^{m-n}} = \frac{2^{n-m}}{2^{n}-1} $$ is in lowest terms. In either case, the denominator in lowest terms is a difference of powers of 2. Convince yourself that of the above answers, only $\frac{1}{7}$ can be represented in this way.

Hint: If $m>n$, then $2^{m-1}\leq2^m-2^n<2^m$

lc2r43
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