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two ways to interpret the same sum

Math newbie here!

The picture above shows my "working out" (or attempt rather). And I've managed to nail down the right name for one of the sums (labeled 1) as a sum of triangle numbers!

But I can't figure out what the second sums is? (if it's even a "complete sum" at all?) It's a bit ugly to use as well (since it requires encoding the 'last digit' of the sum, in order to create a sort of "reverse sum coefficient" that I then multiply by). It was however a much more intuitive sum for me to arrive at, since I started grouping similar numbers. ("here are four ones! and three twos!... etc").

I couldn't really find a good name for this sum (I'm not even quite sure what to google for either ("shearing sum?") as my math vocab is pretty limited!). I keep thinking of the word "shearing" as a good description, since it's like two surfaces going in opposite directions!

yiffyiffyiff
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    Welcome to SE math. Is this sum you're referring to: $\sum_{k=1}^4 k^2 (5-k)$? Side note, the term series is typically reserved for sums with infinite number of terms. – Aaron Hendrickson Dec 26 '19 at 00:32
  • ah right, yeah I'm using the term series wrong . and yeah, that's the sum! the 5 is just 1 greater than the 4 (upper limit), and then I "go backwards" from it by subtracting k! – yiffyiffyiff Dec 26 '19 at 00:48
  • ok, I think I'm using sigma notation wrong here too! (I thought I had to leave a 'k' to show what letter I'm using for the terms of the sigma summation... but now it reads like I'm multiplying by an extra k, sorry to those reading!) – yiffyiffyiff Dec 26 '19 at 01:11
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    I see. As far as a name is concerned, I do not think there is one. Many series (sums in this case) are not formally named and this is probably one of those cases. That said, check out this wikipedia article that talks about sums of this type: https://en.m.wikipedia.org/wiki/Faulhaber%27s_formula?wprov=sfla1 – Aaron Hendrickson Dec 26 '19 at 14:05
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    Also, you may want to do a little reading on sigma notation for sums. You got to learn to crawl before you walk! Note that the index you are summing over is specified in the lower limit under the sigma. In my example, $k$ is used. – Aaron Hendrickson Dec 26 '19 at 14:10

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I don't know whether this sum has a name. We have $$\sum_{k=1}^nk(n+1-k)=(n+1)\cdot\sum_{k=1}^nk-\sum_{k=1}^nk^2=$$ $$=(n+1)\cdot \frac {n(n+1)}{2}-\frac {n(n+1)(2n+1)}{6}=$$ $$=\frac {n(n+1)}{6}\cdot (3(n+1)-(2n+1))=$$ $$=\frac {n(n+1)(n+2)}{6}.$$

DanielWainfleet
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