```Question 137056
Given: The first term of every summation in the series is {{{a}}}.
Given: The number of numbers in each summation is 1 greater than the previous term.
Given: There are {{{n}}} terms in the series.

The first term of your series must be {{{na}}}, because every term in the series will have an {{{a}}} in it.

The second term of your series must be {{{(n-1)(a+1)}}}, because all but the first term will have an {{{a+1}}} in it.  This expands to: {{{na+n-a-1}}}

The third term of your series must be {{{(n-2)(a+2)}}}, similar logic, and this expands to {{{na+2n-2a+4}}}

This gives us the idea that the <i>i</i>-th term must be {{{(n-(i-1))(a+(i-1))}}}, which expands to {{{na+(i-1)n-(i-1)a-(i-1)^2}}}

Notice that every term has an {{{na}}} term, so there must be {{{n}}} times {{{na}}} in your sum.  The first term of the shortcut formula is then {{{n^2a}}}

If you sum all of then {{{n}}} terms, you find that the coefficients can be expressed as {{{sum((i-1),i=1,n)=sum(i,i=0,n)=sum(i,i=1,n-1)}}}.  Since {{{sum(i,i=1,n)=(n(n+1))/2}}}, {{{sum(i,i=1,n-1)=(n(n-1))/2}}}.  Hence, the second term of the shortcut formula must be {{{(n^2(n-1))/2}}}.

Similarly, the sum of all the {{{a}}} terms must be {{{-(na(n-1))/2}}}.

Finally, the last term is {{{-sum((i-1)^2,i=1,n)=-sum(i^2,i=1,n-1)=-(n(n-1)(2n-1))/6}}} because the sum of squares is {{{sum(i^2,i=1,n)=(n(n+1)(2n+1))/6}}}.

Putting it all together:

{{{sum(sum(i,i=1,k),k=1,n)=n^2a+(n^2(n-1))/2-(na(n-1))/2-(n(n-1)(2n-1))/6=n(na+(n-1)(n/2-a/2-(2n-1)/6))}}}

I won't go into the details of the derivation, but if you have a common difference other than 1, call it {{{d}}}, the formula becomes:

{{{sum(sum((i+(i-1)d),i=1,k),k=1,n)=n^2a+(dn^2(n-1))/2-(na(n-1))/2-(dn(n-1)(2n-1))/6=n(na+(n-1)(dn/2-a/2-d(2n-1)/6))}}}```