Question

practice

Answer 1

\[S=\large \sum_{i,j} ij=\sum_1^{40} i\sum_1^{40} j=(\sum _1^{40 }i)^2=(1+2+...40)^2\\=\frac{40(41)}{2}=820^2\]
\[\sum_{i=1}^{40}ij\]