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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
A. 1
B. 2
C. 3
D. π
E. Infinite
maggie27
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
05 Sep 2014, 04:37
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AmoyV
\(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...=1\)
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.
