# Putting into closed form

1. Aug 5, 2008

Hi,

I cant 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.. untill 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

2. Aug 5, 2008

### HallsofIvy

Staff Emeritus
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
$$\frac{5(6)(7)}{6}= (5)(7)= 35$$

Last edited: Aug 5, 2008
3. Aug 5, 2008

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

cheers

4. Aug 5, 2008

### HallsofIvy

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

5. Aug 5, 2008

### HallsofIvy

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

6. Aug 5, 2008

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

7. Aug 6, 2008