Question

Can anyone show me step by step how is this calculated
http://www3.wolframalpha.com/Calculate/MSP/MSP51791a43a081da6hhc92000059de5h65ed9g0ai5?MSPStoreType=image/gif&s=11&w=123&h=51

Answer 1

j = i+1
j = i+2
j = i+3
...
...
j = n-3
j = n-2
j = n-1


There are (n-1)-(i+1)+1 = n-1-i-1+1 = -i+n-1 terms


So you are adding up -i+n-1 terms, ie you are multiplying 1 by -i+n-1 to get the final answer of -i+n-1

Answer 2

The general rule is this

if you start at k and you end at m, where k and m are integers, then you are adding m - k + 1 terms.

Answer 3

What to do if I have something like this:
\[\sum_{i=1}^{n}\sum_{j=i}^{n}\sum_{k=1}^{n-j}1\]

Answer 4

well a double sum means that you have a 2x2 grid of sums (where each row is the complete series of the inner sum)

a triple sum can be thought as a 3d cube (which gets even more complicated)

Answer 5

ie, it's not pretty...

Answer 6

Can you recommend me a tutorial or book which will be useful?
This type of problems occurs in algorithm analysis.

Answer 7

you can try it like this though

[\sum_{i=1}^{n}\sum_{j=i}^{n}\sum_{k=1}^{n-j}1\]

breaks down into

[\sum_{j=i}^{n}\sum_{k=1}^{n-j}1 + \sum_{j=i}^{n}\sum_{k=1}^{n-j}1 + \cdots + \sum_{j=i}^{n}\sum_{k=1}^{n-j}1\]

and there are n copies of \[\sum_{j=i}^{n}\sum_{k=1}^{n-j}1\]

Answer 8

to compute each \[\sum_{j=i}^{n}\sum_{k=1}^{n-j}1\], you can either break it down as shown above or you can make a table