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Aug 5, 2021 at 22:44 vote accept summationman
Aug 5, 2021 at 15:02 comment added Dave L Renfro See the freely available reference in this recent comment of mine.
Aug 5, 2021 at 10:02 comment added Gerry Myerson Thre are books of summation identities, e.g., Jolley, Summation of Series. books.google.com.au/books/about/…
Aug 5, 2021 at 9:09 comment added Wlod AA $\sum_{k=1}^n ka^kb^{n-k}\ =\ b^n\cdot\sum_{k=1}^n\,k\cdot\left(\frac ab\right)^k$
Aug 5, 2021 at 9:08 review Close votes
Aug 8, 2021 at 0:08
Aug 5, 2021 at 8:46 history made wiki Post Made Community Wiki by Stefan Kohl
Aug 5, 2021 at 8:11 answer added Johannes Trost timeline score: 0
Aug 5, 2021 at 8:08 history edited YCor CC BY-SA 4.0
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Aug 5, 2021 at 7:55 comment added summationman Thanks I hadn't thought of that! I did realize I could use the formula I listed by factoring out the $b^n$ term however. If you want to leave ur answer as an answer instead of a comment I will accept it, since it seems pretty general.
Aug 5, 2021 at 7:51 comment added Johannes Trost Try out Wolfram Alpha. The code for your example is wolframalpha.com/input/…
Aug 5, 2021 at 7:47 review First posts
Aug 5, 2021 at 7:57
Aug 5, 2021 at 7:40 history asked summationman CC BY-SA 4.0