∑j=08(j8)(j+1)(j+2)

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$j = 0 \sum 8 (j 8) \frac{1}{( j + 1 ) ( j + 2 )} = 1 \cdot \frac{1}{( 0 + 1 ) ( 0 + 2 )}$

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

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

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

$=\dfrac{1013}{90}$