Generalized formula for sum of products. Q:The sum of all possible products of the first n natural numbers taken two by two   is?
 I did not understand the question as it is.What exactly is being asked?I'd really appreciate an answer explaining what is being asked along with a solution.
 A: It is asked for $\displaystyle\sum\limits_{1\le i<j\le n} i\cdot j$.
Hint: Think of $(1+2+\dots+n)^2$.
A: In my book (where $0\in\Bbb N$) they would probably be asking for
$$
  \sum_{0\leq i<j<n}ij
$$
(but maybe they are negationists who deny $0\in\Bbb N$; then the conditions would be $0<i<j\leq n$ instead). "Taking two by two" usually refers to selecting unordered pairs of distinct elements from a set (here $\{i,j\}$, chosen such that $i<j$ to avoid having the same pair twice); for instance in the Chinese remainder theorem one requires the moduli to be "two by two relatively prime", which means any pair of distinct moduli is required to be relatively prime.
There is (independently of the $0\in\Bbb N$ issue) some ambiguity in their formulation though. Are the numbers $1\times 6,2\times3,3\times 2,6\times 1$ (all of which are equal to $6$), $4$ distinct products, or $2$, or just $1$? I've interpreted them as two products (not counting $2\times3$ and $3\times 2$ as different products), but one could defend there is only one product (namely $6$). The latter interpretation would make the question quite a bit harder (basically one is than taking the sum of all composite numbers up to $(n-1)n$, but including some primes $1\times p$ as well and excluding some squares; rather hard to handle exceptions).
