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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12 [#permalink]
Expert Reply
Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)


21160


\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}} = \frac{1}{2^{10}}*[1+1/2 + 1/4 + 1/8]\)

i.e. \(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}} = \frac{1}{2^{10}}*[15/8] ≈ \frac{1}{2^{9}}\)
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12 [#permalink]
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given
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)
can be written as
1/2^10 * ( 1+1/2+1/4+1/4 )
or say
1/2^10 * 8/4 ; 1/2^10 * 2 ;
1/2^9
OPTION C


Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)


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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12 [#permalink]
Expert Reply
Bunuel wrote:
\(\frac{1}{2^{10}} + \frac{1}{2^{11}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)


A. \(\frac{1}{2^7}\)

B. \(\frac{1}{2^8}\)

C. \(\frac{1}{2^9}\)

D. \(\frac{1}{2^{13}}\)

E. \(\frac{1}{2^{45}}\)




First, let’s obtain the common denominator of 2^12. We obtain:

(2^2/2^2)(1/2^10) + (2/2)1/2^11) + 1/2^12 + 1/2^12

2^2/2^12 + 2/2^12 + 1/2^12 + 1/2^12 = 8/2^12 = 2^3/2^12 = 1/2^9

Answer: C
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12 [#permalink]
Top Contributor
Let try another approach:
The max value in the series is 1/2^10 and the rest of the terms are less than 1/2^10.

since there are 4 terms , we can say that SUM < 4 * 1/2^10
i.e. SUM < 1/2^8

The min value in the series is 1/2^12 and the rest of the terms are greater than 1/2^10.
SUM > 4 * 1/2^12
SUM > 1/2^10


We can conclude that 1/2^10 < SUM < 1/2^8

When we analyze the options, we can see that all options are in decreasing order and only 1/2^9 is in our range.

Option C is the right answer.

Thanks,
Clifin J Francis
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12 [#permalink]
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Re: 1/2^10 + 1/2^11 + 1/2^12 + 1/2^12 [#permalink]
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