Question

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

Answer 1

(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)

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