## Discrete Mathematics with Applications 4th Edition

$\frac{35}{3}$
Starting with i=2, use i to compute the first term. Then continue to move on until you reach i=5. Then add up all the terms you get. $\frac{2\times(2+2)}{(2-1)\times(2+1)}\times\frac{3\times(3+2)}{(3-1)\times(3+1)}\times\frac{4\times(4+2)}{(4-1)\times(4+1)}\times\frac{5\times(5+2)}{(5-1)\times(5+1)}$ $=\frac{8}{3}\times\frac{15}{8}\times\frac{24}{15}\times\frac{35}{24}$ $=\frac{35}{3}$