Answer to Question #275674 in Discrete Mathematics for Zahir

Question

∑j=08​(j8​)(j+1)(j+2)

Answer 1

$$\displaystyle\sum_{j=0}^{8}\dbinom{8}{j}\dfrac{1}{(j+1)(j+2)}=1\cdot\dfrac{1}{(0+1)(0+2)}$$

$$+8\cdot\dfrac{1}{(1+1)(1+2)}+28\cdot\dfrac{1}{(2+1)(2+2)}$$

$$+56\cdot\dfrac{1}{(3+1)(3+2)}+70\cdot\dfrac{1}{(4+1)(4+2)}$$

$$+56\cdot\dfrac{1}{(5+1)(5+2)}+28\cdot\dfrac{1}{(6+1)(6+2)}$$ $$+8\cdot\dfrac{1}{(7+1)(7+2)}+1\cdot\dfrac{1}{(8+1)(8+2)}$$

$$=\dfrac{1013}{90}$$