Compute each of the double double sums below (a)3∑(i=1) 2∑(j=2) (i-j) (b)3∑(i=0) 2∑(j=0) (3i+2j) (c)3∑(i=1) 2∑(j=0) j (d) 2∑(i=0) 3∑(j=0) i^2 j^3

(a)

$\sum_{i=1}^{3}\sum_{j=2}^{2}( i-j) =\sum_{i=1}^{3}(i-2)=1-2+2-2+3-2=0$

(b)

$\sum_{i=0}^{3} \sum_{j=0}^{2} (3i+2j) = \sum_{i=0}^{3}( 3i+(3i+2)+(3i+4)) \newline = \sum_{i=0}^{3}( 9i+6) =6+ 9+6+9\cdot2+6+9\cdot3+6=78$

(c)

$\sum_{i=1}^{3} \sum_{j=0}^{2} j = \sum_{i=1}^{3}( 0+1+2)=3\cdot3=9$

(d)

$\sum_{i=0}^{2} \sum_{j=0}^{3} i^2 j^3 = \sum_{i=0}^{2}( i^2 + 8i^2 +27i^2) \newline = \sum_{i=0}^{2}36i^2 = 36 + 36 \cdot4=36\cdot5=180$