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After reducing each term we get 4*1/(2^12)= 1/(2^10)
IMO A

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Bunuel
\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16

simplify expression
we get ; 4/2^12 ; 1/2^10
IMO A
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Bunuel
\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16

\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}}\)

\(=\frac{1}{2^{12}} + \frac{1}{2^{12}} + \frac{1}{2^{12}} + \frac{1}{2^{12}}\)

\(=\frac{4}{2^{12}}\)

\(=\frac{1}{2^{10}}\); Answer must be (A)
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Bunuel
\(\frac{1}{2^{12}} + \frac{2}{2^{13}} + \frac{4}{2^{14}} + \frac{8}{2^{15}} =\)


A. 1/2^10

B. 1/2^12

C. 15/2^15

D. 2/2^10

E. 23/2^16

I would say, simply take 1/2^12 common.

--> 1/2^12 [1+2/2+4/4+8/8] = 1/2^12[4] = 1/2^2 * 2^10 [4] = 1/2^10 cheers
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