sum of the series: sum [k=1, n] 2^{n-k} k^2

sarathys

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Sep 13, 2017
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Evaluate the sum:

SIGMA i=1 to n 2n-i . i2
 
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Evaluate the sum of the series

\(\displaystyle \displaystyle \sum_{k=1}^n \: 2^{n-k} \cdot k^2\)


What are your thoughts? What have you tried? For instance, you split apart the \(\displaystyle 2^{n-k}\) term into \(\displaystyle 2^n \cdot 2^{-k}\), then recognized that \(\displaystyle 2^n\) is a constant so you pulled it out front of the sum... and then what? Please re-read the Read Before Postinghttps://www.freemathhelp.com/forum/threads/41536-Read-Before-Posting!! thread that's stickied at the top of every subforum, and comply with the rules contained within. Please share with us any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.
 
sum of the series [k=1,n] 2^(n-k).k^2

Yes I got it simplified as

S= 2^n[ (1/2)+(2^2/2^2)+(3^2/2^3)+(4^2/2^4)+............+(n^2/2^n)]

Now I do not know how to find the sum to n terms of the series specified in [ ]
 
According to Wolfram Alpha, the answer is supposed to be:

. . . . .\(\displaystyle -n^2\, -\, 4n\, +\, 6\, (2^n\, -\, 1)\)

Off the top of my head, I have no idea how one is expected to figure this out...? :shock:
 
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