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