How can I put this summation into a closed form?

[beadmaster]

Hi,

I can't think / remeber how to write the following expresion in a closed form,

the function is a summation of natural numbers between 1 and an upper limit "K", written as

Sigma x with limits K and 1 effectivly, straightforward etc...
what i want is the summation of all the "summations" between K and 1 so sigma (1,K) + sigma (1,(K-1)) + sigma (1,(K-2)) etc.. until it reaches sigma (1,1) ie 1.

its easy to visulise, take K as 5, the expression would be

5+4+3+2+1
+4+3+2+1
+3+2+1
+2+1
+1

which gives 35, i am wanting the answer in terms of K (im presuming its possible) or at least can be written much neater than an expansion, cheers (I thought it could possibly be written as a general sigma summation but prehaps with tending limits, but if so how do you represent the tending being discrete and not continous)

cheers,
tom

 

[HallsofIvy]

It is fairly well known that
\sum_{i=1}^n i= \frac{n(n+1)}{2}
so your "sum of sums" is
\sum_{k=1}^n\left(\sum_{i=1}^k i\right)= \sum_{k=1}^n \frac{k(k+1)}{2}

It is also (reasonably) well known that
\sum_{k=1}^n k^2= \frac{n(n+1)(2n+1)}{6}

putting those together,
\sum_{k=1}^n \frac{k(k+1)}{2}= \frac{1}{2}\sum_{k=1}^n k^2+ \frac{1}{2}\sum_{k=1}^n k
= \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2}\right)= \frac{1}{4}n(n+1)\left(\frac{2n+1}{3}+ \frac{3}{3}\right)= \frac{n(n+1)(n+2)}{6}

You will note that for your example, with n= 5, this gives
\frac{5(6)(7)}{6}= (5)(7)= 35

 

[beadmaster]

thats great thanks, i knew the general summations but tbh i wouldn't have the wit to make some of those connections, not currently at least...:P

cheers

 

[HallsofIvy]

That turns out be remarkably simple! I hadn't expected it to.

 

[HallsofIvy]

Actually, I was surprised that the final formula was so simple!

 

[beadmaster]

i recognise the final answer from somewhere else, o well thanks again
:)

 

[beadmaster]

can i just mention that that "proof" or process appears increadibly elegant, how long did it take for you to do, could i be so intrusive to ask about you level of expertise as i am taken back by again by the elegence of the solution

cheers