MHB Sum of Products of N Natural Numbers

[mathworker]

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]

Re: sum of products of N

i tried with following approach ,
S=$$\sum n^2 + \sum n(n-1)...\sum n(n-(n-1))$$...
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]

Re: sum of products of N

i tried with following approach ,
S=$$\sum n^2 + \sum n(n-1)...\sum n(n-(n-1))$$...
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]

yeah i got from Opalgs idea,
$$Sn - S(n-1)=n(n(n+1))/2$$
$$S(n-1)-S(n-2)=(n-1)(n-1)n/2[/math]
and by canceling up to $$S_0$$
$$Sn = \sum n^2(n+1)/2$$
$$=n(n+1)(n+2)(3n+1)/24$$
thanks oplag...:D