Question 892087
Arrange the sums in this manner:

1 + 2 + 3 + 4 + ... + n + (n+1) + ... + x
    2 + 3 + 4 + ... + n + (n+1) + ... + x
        3 + 4 + ... + n + (n+1) + ... + x
            4 + ... + n + (n+1) + ... + x
                     ...
                (n-1)+n + (n+1) + ... + x
                      n + (n+1) + ... + x
            
The triangular portion of the sums can be expressed as 
1 + 2*2 + 3*3 + 4*4 +... + (n-1)*(n-1) + n*n, or 
{{{1+2^2 + 3^2 +4^2}}}+...+{{{(n-1)^2 +n^2 = (n*(n+1)*(2n+1))/6}}}
Now the sum (n+1)+...+x is added n times.  Hence
the sum {{{n*((x-n)/2)*(n+1+x)}}}, by using the sum of an arithmetic sequence.

Therefore, (1+2+3....+x) + (2+3+4....+x) + (3+4+5....+x) + .... till n terms where n < x, is
{{{(n*(n+1)*(2n+1))/6 + (n*(x-n)*(n+1+x))/2}}}