(1 + 1/2)(1 + 1/4)...(1 + 1/2^32) = ?

Question

(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) = ?

A.  1- \frac{1}{2^{64}}

B.  2(1- \frac{1}{2^{64}})

C.  4(1- \frac{1}{2^{32}})

D.  4(1+ \frac{1}{2^{32}})

E.  \frac{1}{2^{64}}

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ThatDudeKnows · Accepted Answer

Why would I work with fractions when I don't need to? And why would I do the real math when I can ballpark my way to a right answer?

(1.5)*(1.25)*(the last four elements are REALLY close to 1...can we please just agree that they're 1...or maybe somewhere between 1 and 1.1 if you feel better about that?)

~(1.9)*~(1.05)

Looks like right around 2.

Let's ballpark the answer choices.

A) 1
B) 2
C) a teeeeny bit less than 4
D) a teeeeny bit more than 4
E) a teeeeeeeeeeeny bit more than 0

Answer choice B.

ThatDudeKnowsBallparking

false9 · Answer

Call the expression X.
X = (1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)(1+1/2^32)

Multiply by (1-1/2) in the numerator and denominator,

X = [(1-1/2)(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)(1+1/2^32)]/(1-1/2)

X = [(1-1/2^2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16)(1+1/2^32)]/(1-1/2)

after solving this, we will have;

X = [(1-1/2^64)]/(1-1/2)
X= 2(1-1/2^64). Option B.

Feedbacks are welcome.

Kinshook · Answer

Asked:  (1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) = ?

(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) =(1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) / (1 - \frac{1}{2}) = 2(1-\frac{1}{2^{64}})

IMO B

Regor60 · Answer

How many powers of 2 in the denominator ? Easy to count.

1+2+4+8+16+32 = 63

There are no powers of 2 in the numerator because each numerator is odd due to the addition of 1, so the answer must have 2^63 in the denominator.

Only answer B satisfies

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cat2010 · Answer

Formula :  (a-b)(a+b)=  a^2-b^2

(1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}})

Multiply and divide by  (1-\frac{1}{2})

= (1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) / (1 - \frac{1}{2})

Denominoter will be \frac{ 1}{2} as (1-\frac{1}{2}) = \frac{1}{2} 
When it goes up in numerator it will be 2

= 2(1-\frac{1}{2})(1+\frac{1}{2})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) 
 
Now , (1-\frac{1}{2})(1+\frac{1}{2}) = (1-\frac{1}{2^{2}}) ....As  (a-b)(a+b)= a^2-b^2

So ,our expression will become 
= 2(1-\frac{1}{2^{2}})(1+\frac{1}{2^2})(1+\frac{1}{2^4})...(1+\frac{1}{2^{32}}) 
 
Again ,  (1-\frac{1}{2^{2}})(1+\frac{1}{2^2}) =(1-\frac{1}{2^{4}}) ....As   (a-b)(a+b)=a^2-b^2

So ,our expression will become 
= 2(1-\frac{1}{2^{4}})(1+\frac{1}{2^4})(1+\frac{1}{2^8})...(1+\frac{1}{2^{32}})

See the pattern ....You will easily get that the aswer will B as

So ,our expression will become last 
= 2(1-\frac{1}{2^{64}}))

IMO B

bumpbot · Answer

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(1 + 1/2)(1 + 1/4)...(1 + 1/2^32) = ?

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(1 + 1/2)(1 + 1/4)...(1 + 1/2^32) = ?

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