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If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is

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If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post Updated on: 02 Aug 2014, 14:57
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If the infinite sum \(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...=1\), what is the value of the infinite sum \(\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+....\)?


A. 1

B. 2

C. 3

D. π

E. Infinite

Originally posted by maggie27 on 23 Jul 2014, 22:35.
Last edited by maggie27 on 02 Aug 2014, 14:57, edited 1 time in total.
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Re: If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post 24 Jul 2014, 00:04
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maggie27 wrote:
If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is the value of the infinite sum 1/2^1+2/2^22+3/2^3+4/2^4+....?

A) 1
B) 2
C) 3
D) π
E) Infinite


Sum \(= 1/2^1 + 1/2^2 + 1/2^3+1/2^4+... = 1\)
Sum = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 ..............(I)

Note that the first term is 1/2 so the sum of the rest of the terms must be 1/2 to add up to a total of 1.
So 1/4 + 1/8 + 1/16 + 1/32 + ... = 1/2

Similarly, now the first term is 1/4 so the sum of the rest of the terms must be 1/4 too to get a sum of 1/2.

and so on...

Required sum \(= \frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+....\)

Required sum \(=(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^3}+\frac{1}{2^4}+...) + ...\)

Required sum \(= (1) + (\frac{1}{2}) + (\frac{1}{4}) + (\frac{1}{8})+ ...\)

From equation (I) above, we know that 1/2 + 1/4 + 1/8 + ... = 1
So Required sum = 1 + 1 = 2
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Re: If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post 24 Jul 2014, 00:08
maggie27 wrote:
If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is the value of the infinite sum 1/2^1+2/2^22+3/2^3+4/2^4+....?

A) 1
B) 2
C) 3
D) π
E) Infinite


A crude method


1/2^1+2/2^22+3/2^3+4/2^4+....n/2^n

1/2+1/2+3/8+4/16+5/32+6/64+7/128+8/512+...

1+0.375+0.25+0.15+0.09+0.04+0.015... <2

So infinite sum should equal 2 as n increases 2^n increases exponentially.
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Re: If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post Updated on: 05 Sep 2014, 04:39
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If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is the value of the infinite sum 1/2^1+2/2^2+3/2^3+4/2^4+....?
A. 1
B. 2
C. 3
D. π
E. Infinite

Intuitive, short and uncomplicated explanation please.

Originally posted by AmoyV on 05 Sep 2014, 04:10.
Last edited by AmoyV on 05 Sep 2014, 04:39, edited 1 time in total.
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Re: If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post 05 Sep 2014, 04:37
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1
AmoyV wrote:
If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is the value of the infinite sum 1/2^1+2/2^2+3/2^3+4/2^4+....?
A. 1
B. 2
C. 3
D. π
E. Infinite

\(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...=1\)

Now, \(\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...\) can be written as

\((\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+...) + (\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+...) +...\)

\(= 1+ [(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) - \frac{1}{2^1}] + [(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) - \frac{1}{2^1} - \frac{1}{2^2}] +...\)

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

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

So, the answer is B.

Hope that helps.
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Re: If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post 05 Sep 2014, 07:03
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1
Let
S= 1/2+2/2^2+3/2^3.......

- S/2= 1/2^2+2/2^3........ dividing both sides by half and subtracting)

S/2 = 1/2+1/2^2+1/2^3.....(This R.H.S is given to be 1 in question)

So
S/2=1
S=2
Hence the answer
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Re: If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post 07 Apr 2018, 02:11
VeritasPrepKarishma wrote:
maggie27 wrote:
If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is the value of the infinite sum 1/2^1+2/2^22+3/2^3+4/2^4+....?

A) 1
B) 2
C) 3
D) π
E) Infinite


Sum \(= 1/2^1 + 1/2^2 + 1/2^3+1/2^4+... = 1\)
Sum = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 ..............(I)

Note that the first term is 1/2 so the sum of the rest of the terms must be 1/2 to add up to a total of 1.
So 1/4 + 1/8 + 1/16 + 1/32 + ... = 1/2

Similarly, now the first term is 1/4 so the sum of the rest of the terms must be 1/4 too to get a sum of 1/2.

and so on...

Required sum \(= \frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+....\)

Required sum \(=(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...) + (\frac{1}{2^3}+\frac{1}{2^4}+...) + ...\)

Required sum \(= (1) + (\frac{1}{2}) + (\frac{1}{4}) + (\frac{1}{8})+ ...\)

From equation (I) above, we know that 1/2 + 1/4 + 1/8 + ... = 1
So Required sum = 1 + 1 = 2

Hi, Karishma!!
Can you, please, elaborate how did you derive the highlighted line?
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If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is  [#permalink]

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New post 07 Apr 2018, 14:22
1/2 = .5
2/4 = 1/2 = .5
3/8 = .375
1/4 = .25

Sum together = 1.58..

The value is approaching 2.

Answer B
If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is &nbs [#permalink] 07 Apr 2018, 14:22
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