06/24/2010, 06:53 PM
There exists a formula for iterated sums.
If
\( Sum_n({ a_n }) = S_n \)
\( S_0 = 0 \)
\( S_n = S_{n - 1} + a_n \)
Then
\( Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j \)
If
\( Sum_n({ a_n }) = S_n \)
\( S_0 = 0 \)
\( S_n = S_{n - 1} + a_n \)
Then
\( Sum_n^k({ a_n }) = \sum_{j=1}^{n} { {k + (n - j) - 1}\choose{k - 1} } \cdot a_j \)