Evaluating a double summation

First fix $i$ and evaluate the inner sum over $j$: $\sum_{j=1}^{3} (i\cdot j) = i(1+2+3) = 6i$, since $i$ is treated as a constant while summing over $j$.

Now substitute this result into the outer sum: $\sum_{i=1}^{4} 6i = 6\sum_{i=1}^{4} i = 6(1+2+3+4) = 6(10) = 60$. So the double sum equals $60$.