2阶\[\sum_{i=l}^{r}\sum^{i}_{j=1}a_j\]\[=\sum_{i=l}^{r}(r-i+1)a_i\]\[=(r+1)\sum_{i=l}^{r}a_i+\sum_{i=l}^{r}i\cdota_i\]这个很好理解，因为对于第\(i\)个数，他加了\((r-i+1)\)次三阶正常拆解\[\sum_{i=l}^{r}\sum^{i}_{j=1}\sum_{k}^{j}a_k\]\[=