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

Question

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

KarishmaB · Accepted Answer

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 Another quirky series question is discussed here: https://youtu.be/KX8WNiyNUIo

eshan429 · Answer

\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+   +   +...

= 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.

WoundedTiger · Answer

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.

AmoyV · Answer

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.

utkgogia2003 · Answer

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

NaeemHasan · Answer

Hi, Karishma!!
Can you, please, elaborate how did you derive the highlighted line?

Funsho84 · Answer

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

NHasan19058 · Answer

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

Rakesh1987 · Answer

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

rakeshmishra5690 · Answer

given, 1/2 + 1/2^2 +1/2^3 + ...................... = 1

Let S = 1/2 + 2/2^2 + 3/2^3 ............................. (1)
 S = 1/2^1 +2/2^2 ................................(2)

multiply (2) with 1/2 and subtract to (1)

1/2*S = 1/2 + 1/2^2 +1/2^3 + ...................... 
So S = 2

Ans B

Goldenfuture · Answer

She has basically made a GP and sum of infinite GP is a/1-r
= 1/ 1/2 = 2

luisdicampo · Answer

Deconstructing the Question  
Given:  \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 .
Target: Find the value of  S = \sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \dots

Method: Arithmetico-Geometric Series Manipulation  
Let the sum be  S :
 S = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \dots  (Equation 1)

Multiply the entire sum by the common ratio  \frac{1}{2} :
 \frac{1}{2}S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots  (Equation 2)

Subtract Equation 2 from Equation 1:
 S - \frac{1}{2}S = \frac{1}{2} + (\frac{2}{4} - \frac{1}{4}) + (\frac{3}{8} - \frac{2}{8}) + (\frac{4}{16} - \frac{3}{16}) + \dots

\frac{1}{2}S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots

The Right Hand Side is a standard infinite geometric series with  a=\frac{1}{2}  and  r=\frac{1}{2} .
The problem statement explicitly tells us that this sum equals  1 .

So:
 \frac{1}{2}S = 1 
 S = 2

Answer:  B

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

Prep Toolkit

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My Rewards