text: 1. compute a sum
  
  hello
  
  i need to compute this sum:
  
  sigma(n=1 to infinity) n^2 / 2^n
  
  and i'm having trouble doing that.
  i tried looking at a power series,but couldn't develop it.
  
  help would be appreciated.
  thanks.
  
  2. Originally Posted by parallel
  hello
  
  i need to compute this sum:
  
  sigma(n=1 to infinity) n^2 / 2^n
  
  and i'm having trouble doing that.
  i tried looking at a power series,but couldn't develop it.
  
  help would be appreciated.
  thanks.
  This is a bit cryptic, but if you can do it from this you will have done
  a lot of the work yourself. If you have trouble we will fill in the detail.
  
  Consider the geometric series:
  
  $\displaystyle S(x)=\sum_1^{\infty} (x/2)^n$
  
  which you should be able to sum for $\displaystyle |x|<2$. Now differentiate
  once and twice and see what you find after a bit of rearrangement of the
  corresponding sums of the series, and letting $\displaystyle x=1$.
  
  RonL
  
  3. Hello, parallel!
  
  I just "invented" an approach to this problem . . .
  
  $\displaystyle \sum^{\infty}_{n=1}\frac{n^2}{2^n}$
  
  We have: .$\displaystyle S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots$ [1]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots$ [2]
  
  Subtract [2] from [1]:
  
  . . . . . . . $\displaystyle \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots$ [3]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots$ [4]
  
  Subtract [4] from [3]:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots$
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$
  . . . . . . . . . . . . . . an infinite geometric series with $\displaystyle a = \frac{1}{2},\;r = \frac{1}{2}$
  . . . . . . . . . . . . . . . . . . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$
  
  Hence, we have: .$\displaystyle \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2}$ . . . . (drumroll, please)
  
  Therefore: .$\displaystyle \boxed{S\;=\;6}$
  . . . . ta-DAA!
  
  4. Originally Posted by Soroban
  Hello, parallel!
  
  I just "invented" an approach to this problem . . .
  
  We have: .$\displaystyle S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots$ [1]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots$ [2]
  
  Subtract [2] from [1]:
  
  . . . . . . . $\displaystyle \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots$ [3]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots$ [4]
  
  Subtract [4] from [3]:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots$
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$
  . . . . . . . . . . . . . . an infinite geometric series with $\displaystyle a = \frac{1}{2},\;r = \frac{1}{2}$
  . . . . . . . . . . . . . . . . . . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$
  
  Hence, we have: .$\displaystyle \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2}$ . . . . (drumroll, please)
  
  Therefore: .$\displaystyle \boxed{S\;=\;6}$
  . . . . ta-DAA!
  
  This is implicit in the method I outlined
  
  CB
  
  5. thanks for the reply CaptainBlack,this is what I came up with,hope it's correct
  
  the sum of (x/2)^n is : x/(2-x)
  
  taking deriv:
  (1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2
  
  taking 2nd deriv:
  
  sigma n(n-1)/2^n * x^(n-2)
  this sum is:
  
  (2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4
  
  now I noticed that the second term is (1) * (1/x),
  so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)
  
  is this correct?
  thanks for the help
  
  Soroban,
  hey, I really liked your solution,I'll sure keep this approach in mind
  
  6. Originally Posted by parallel
  thanks for the reply CaptainBlack,this is what I came up with,hope it's correct
  
  the sum of (x/2)^n is : x/(2-x)
  
  taking deriv:
  (1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2
  
  taking 2nd deriv:
  
  sigma n(n-1)/2^n * x^(n-2)
  this sum is:
  
  (2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4
  
  now I noticed that the second term is (1) * (1/x),
  so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)
  
  is this correct?
  thanks for the help
  
  Soroban,
  hey, I really liked your solution,I'll sure keep this approach in mind
  That's what I had in mind. I will not check it in detail as the answer agrees
  with Soroban's so we may presume the algebra is correct.
  
  This method is powerful as it provides a method of finding:
  
  $\displaystyle S=\sum_1^{\infty}P(n)/k^n$
  
  for any polynomial $\displaystyle P(n)$ in $\displaystyle n$ (at least for
  smallish polynomials) and $\displaystyle k>1$.
  
  RonL
  
  7. thanks again
explanation: The document contains a math question and multiple answers to this question. The question is about computing the sum of a series, and multiple users provide different approaches to solving it.
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distilabel_metadata.content: 1. compute a sum
  
  hello
  
  i need to compute this sum:
  
  sigma(n=1 to infinity) n^2 / 2^n
  
  and i'm having trouble doing that.
  i tried looking at a power series,but couldn't develop it.
  
  help would be appreciated.
  thanks.
  
  2. Originally Posted by parallel
  hello
  
  i need to compute this sum:
  
  sigma(n=1 to infinity) n^2 / 2^n
  
  and i'm having trouble doing that.
  i tried looking at a power series,but couldn't develop it.
  
  help would be appreciated.
  thanks.
  This is a bit cryptic, but if you can do it from this you will have done
  a lot of the work yourself. If you have trouble we will fill in the detail.
  
  Consider the geometric series:
  
  $\displaystyle S(x)=\sum_1^{\infty} (x/2)^n$
  
  which you should be able to sum for $\displaystyle |x|<2$. Now differentiate
  once and twice and see what you find after a bit of rearrangement of the
  corresponding sums of the series, and letting $\displaystyle x=1$.
  
  RonL
  
  3. Hello, parallel!
  
  I just "invented" an approach to this problem . . .
  
  $\displaystyle \sum^{\infty}_{n=1}\frac{n^2}{2^n}$
  
  We have: .$\displaystyle S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots$ [1]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots$ [2]
  
  Subtract [2] from [1]:
  
  . . . . . . . $\displaystyle \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots$ [3]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots$ [4]
  
  Subtract [4] from [3]:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots$
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$
  . . . . . . . . . . . . . . an infinite geometric series with $\displaystyle a = \frac{1}{2},\;r = \frac{1}{2}$
  . . . . . . . . . . . . . . . . . . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$
  
  Hence, we have: .$\displaystyle \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2}$ . . . . (drumroll, please)
  
  Therefore: .$\displaystyle \boxed{S\;=\;6}$
  . . . . ta-DAA!
  
  4. Originally Posted by Soroban
  Hello, parallel!
  
  I just "invented" an approach to this problem . . .
  
  We have: .$\displaystyle S\;=\;\frac{1}{2} + \frac{4}{2^2} + \frac{9}{2^3} + \frac{16}{2^4} + \frac{25}{32} + \hdots$ [1]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{2}S \;= \qquad\frac{1}{2^2} + \frac{4}{2^3} + \frac{9}{2^4} + \frac{16}{2^5} + \hdots$ [2]
  
  Subtract [2] from [1]:
  
  . . . . . . . $\displaystyle \frac{1}{2}S\;=\;\frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \frac{7}{2^4} + \frac{9}{2^5} + \hdots$ [3]
  
  Multiply by $\displaystyle \frac{1}{2}$:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;= \qquad\frac{1}{2^2} + \frac{3}{2^3} + \frac{5}{2^4} + \frac{7}{2^5} + \hdots$ [4]
  
  Subtract [4] from [3]:
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{2}{2^3} + \frac{2}{2^4} + \frac{2}{2^5} + \hdots$
  
  . . . . . . . $\displaystyle \frac{1}{4}S\;=\;\frac{1}{2} + \underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}$
  . . . . . . . . . . . . . . an infinite geometric series with $\displaystyle a = \frac{1}{2},\;r = \frac{1}{2}$
  . . . . . . . . . . . . . . . . . . Its sum is: .$\displaystyle \frac{\frac{1}{2}}{1 - \frac{1}{2}} \:= \:1$
  
  Hence, we have: .$\displaystyle \frac{1}{4}S \;= \;1 + \frac{1}{2} \;= \;\frac{3}{2}$ . . . . (drumroll, please)
  
  Therefore: .$\displaystyle \boxed{S\;=\;6}$
  . . . . ta-DAA!
  
  This is implicit in the method I outlined
  
  CB
  
  5. thanks for the reply CaptainBlack,this is what I came up with,hope it's correct
  
  the sum of (x/2)^n is : x/(2-x)
  
  taking deriv:
  (1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2
  
  taking 2nd deriv:
  
  sigma n(n-1)/2^n * x^(n-2)
  this sum is:
  
  (2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4
  
  now I noticed that the second term is (1) * (1/x),
  so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)
  
  is this correct?
  thanks for the help
  
  Soroban,
  hey, I really liked your solution,I'll sure keep this approach in mind
  
  6. Originally Posted by parallel
  thanks for the reply CaptainBlack,this is what I came up with,hope it's correct
  
  the sum of (x/2)^n is : x/(2-x)
  
  taking deriv:
  (1) sigma n/(2^n)*x^(n-1) = 2/(2-x)^2
  
  taking 2nd deriv:
  
  sigma n(n-1)/2^n * x^(n-2)
  this sum is:
  
  (2) n^2/2^n*x^(n-2) - n/2^n*x^(n-2) = deriv of (1) = (8-4x)/(2-x)^4
  
  now I noticed that the second term is (1) * (1/x),
  so (2) + (1)*(1/x) yields the result,which is (4x+4x-2x^2)/(2-x)^3 = 6 (if x=1)
  
  is this correct?
  thanks for the help
  
  Soroban,
  hey, I really liked your solution,I'll sure keep this approach in mind
  That's what I had in mind. I will not check it in detail as the answer agrees
  with Soroban's so we may presume the algebra is correct.
  
  This method is powerful as it provides a method of finding:
  
  $\displaystyle S=\sum_1^{\infty}P(n)/k^n$
  
  for any polynomial $\displaystyle P(n)$ in $\displaystyle n$ (at least for
  smallish polynomials) and $\displaystyle k>1$.
  
  RonL
  
  7. thanks again
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and multiple answers to this question. The question is about computing the sum of a series, and multiple users provide different approaches to solving it.
  
  Label: question-and-answer