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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
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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


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
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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 May 2019, 01:06
eshan429 wrote:
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.

How have you got the third line as you have written can be written as? I want to understand how the third line is derived
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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 May 2019, 02:35
1
NHasan19058 wrote:
eshan429 wrote:
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.

How have you got the third line as you have written can be written as? I want to understand how the third line is derived


Hi NHasan19058,

Let me explain how the third line is derived.

We need to find the sum of the series given below:
\(\frac{1}{2^1}\)+\(\frac{2}{2^2}\)+\(\frac{3}{2^3}\)+\(\frac{4}{2^4}\)+...

In the next line we will make all numerators 1 and will group all the fractions in decreasing order. As the fist term \(\frac{1}{2^1}\) has numerator 1, therefore it occurs only once in the third line (will be only in first bracket). The second term \(\frac{2}{2^2}\) has numerator 2 i.e. it can be expresses as \(\frac{1}{2^2} +\frac{1}{2^2}\) (will be there in two lists first and second), the third term with numerator 3 will be there in three brackets and so on. Thus grouping together we get:
{\(\frac{1}{2^1}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+\(\frac{1}{2^4}\)+...}+{\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+\(\frac{1}{2^4}\)+\(\frac{1}{2^5}\)...}+{\(\frac{1}{2^3}\)+\(\frac{1}{2^4}\)+\(\frac{1}{2^5}\)+\(\frac{1}{2^6}\)...}

Each of these brackets contains a series which resembles the given series which sums to 1 but starting at different terms-- first series starts with the first term of original series, second series starts with second term of original series, third series starts with third term of original series and so on.
={1}+{1-\(\frac{1}{2}\)}+{1-\(\frac{1}{2}\)-\(\frac{1}{4}\)}+...
=1+\(\frac{1}{2}\)+\(\frac{1}{4}\)+...
=1+1
=2

Please hit +1 Kudos if you liked the explanation.
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If the infinite sum 1/2^1+1/2^2+1/2^3+1/2^4+...=1, what is   [#permalink] 07 May 2019, 02:35
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