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=\(\displaystyle \sum n^2 + \sum n(n-1)...\sum n(n-(n-1))\)...
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]

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]

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

 

[blue_raver22]

The general formula for the sum of products of n natural numbers can be written as follows:

S = (n-1)*(n-2)/2 + (n-1)*(n-2)*(n-3)/6 + ... + (n-1)*(n-2)*(n-3)*...*2*1

This formula can be derived by considering the pattern in the given example. For a set of n natural numbers, the first term in the sum will be the product of the first two numbers, the second term will be the product of the first three numbers, and so on. This pattern can be continued until the last term, which will be the product of all n numbers.

To find the general formula, we can use the concept of combinations. The first term in the sum represents the number of ways we can choose 2 numbers from a set of n numbers. The second term represents the number of ways we can choose 3 numbers from a set of n numbers, and so on. This can be written as nC2, nC3, ..., nCn. Therefore, the general formula can be written as:

S = nC2 + nC3 + ... + nCn

Using the formula for combinations, nCk = n! / (k!(n-k)!), we can simplify the above formula to:

S = n! * (1/2 + 1/6 + ... + 1/n!)

This formula can also be written as:

S = n! * (1/2! + 1/3! + ... + 1/n!)

Or, in a more compact form:

S = n! * ∑(1/k!)

Therefore, for any given n-tuple containing consecutive natural numbers, the sum of products of all possible pairs of numbers can be calculated using this general formula.