SOLUTION: Is there a short way to do the SUM of a sequence of sums: (1) + (1+2) + (1+2+3) + (1+2+3+4) + .... (1+2+3+4....+100). I know how to do the sum of the sequence (i.e. sum of 100= 100

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Question 137056: Is there a short way to do the SUM of a sequence of sums: (1) + (1+2) + (1+2+3) + (1+2+3+4) + .... (1+2+3+4....+100). I know how to do the sum of the sequence (i.e. sum of 100= 100/2(100 + 1)) and I can get the answer, I am just trying to see if there is an easier way then adding each number up. Like, sum for 100 is 5050; sum of 99 is sum of 100 minus 100 or 4050; but I can't find a shortcut and I can't find a formula anywhere for the sum of a series of sums. Can someone help? This is actually not for a class.
Answer by solver91311(12126) About Me  (Show Source):
You can put this solution on YOUR website!
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 %28n-1%29%28a%2B1%29, because all but the first term will have an a%2B1 in it. This expands to: na%2Bn-a-1

The third term of your series must be %28n-2%29%28a%2B2%29, similar logic, and this expands to na%2B2n-2a%2B4

This gives us the idea that the i-th term must be %28n-%28i-1%29%29%28a%2B%28i-1%29%29, which expands to na%2B%28i-1%29n-%28i-1%29a-%28i-1%29%5E2

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%5E2a

If you sum all of then n terms, you find that the coefficients can be expressed as sum%28%28i-1%29%2Ci=1%2Cn%29=sum%28i%2Ci=0%2Cn%29=sum%28i%2Ci=1%2Cn-1%29. Since sum%28i%2Ci=1%2Cn%29=%28n%28n%2B1%29%29%2F2, sum%28i%2Ci=1%2Cn-1%29=%28n%28n-1%29%29%2F2. Hence, the second term of the shortcut formula must be %28n%5E2%28n-1%29%29%2F2.

Similarly, the sum of all the a terms must be -%28na%28n-1%29%29%2F2.

Finally, the last term is -sum%28%28i-1%29%5E2%2Ci=1%2Cn%29=-sum%28i%5E2%2Ci=1%2Cn-1%29=-%28n%28n-1%29%282n-1%29%29%2F6 because the sum of squares is sum%28i%5E2%2Ci=1%2Cn%29=%28n%28n%2B1%29%282n%2B1%29%29%2F6.

Putting it all together:
sum%28sum%28i%2Ci=1%2Ck%29%2Ck=1%2Cn%29=n%5E2a%2B%28n%5E2%28n-1%29%29%2F2-%28na%28n-1%29%29%2F2-%28n%28n-1%29%282n-1%29%29%2F6=n%28na%2B%28n-1%29%28n%2F2-a%2F2-%282n-1%29%2F6%29%29

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%28sum%28%28i%2B%28i-1%29d%29%2Ci=1%2Ck%29%2Ck=1%2Cn%29=n%5E2a%2B%28dn%5E2%28n-1%29%29%2F2-%28na%28n-1%29%29%2F2-%28dn%28n-1%29%282n-1%29%29%2F6=n%28na%2B%28n-1%29%28dn%2F2-a%2F2-d%282n-1%29%2F6%29%29