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Sum of products of N

mathworker

Well-known member
May 31, 2013
119
If we are given with set of n natural numbers and asked to find the sum of all possible products of two number..
E.G:
{1,2,3} is given then ,
1*1+1*2+1*3+2*2+2*3+3*3=s
how to find the general formula of SUM for given n-tuple containing consecutive natural numbers not necessarily starts with 1 or at least first N natural numbers
 

mathworker

Well-known member
May 31, 2013
119
Re: sum of products of N

i tried with following approach ,
S=\(\displaystyle \sum n^2 + \sum n(n-1).......\sum n(n-(n-1))\)...
and tried finding sum by splitting terms but it gone wrong 'cause there are some negative terms which are meant to be ignored(Lipssealed)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,786
Re: sum of products of N

i tried with following approach ,
S=\(\displaystyle \sum n^2 + \sum n(n-1).......\sum n(n-(n-1))\)...
and tried finding sum by splitting terms but it gone wrong 'cause there are some negative terms which are meant to be ignored(Lipssealed)
You should be able to check that the formula for $S_n$ is $S_n = \frac1{24}n(n+1)(n+2)(3n+1)$. Prove this by induction, noting that to get from $S_n$ to $S_{n+1}$ you need to add the terms $(n+1)(1+2+\ldots+(n+1)).$
 

mathworker

Well-known member
May 31, 2013
119
yeah i got from Opalgs idea,
\(\displaystyle Sn - S(n-1)=n(n(n+1))/2\)
\(\displaystyle S(n-1)-S(n-2)=(n-1)(n-1)n/2\)
and by canceling up to \(\displaystyle S_0\)
\(\displaystyle Sn = \sum n^2(n+1)/2\)
\(\displaystyle =n(n+1)(n+2)(3n+1)/24\)
thanks oplag...:D