Solution to j=0 ∑ 8 ​ ( j 8 ​ ) (j+1)(j+2) 1 ​ =1⋅ (0+1)(0+2) 1 … - Sikademy
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j=0 ∑ 8 ​ ( j 8 ​ ) (j+1)(j+2) 1 ​ =1⋅ (0+1)(0+2) 1 ​ +8\cdot\dfrac{1}{(1+1)(1+2)}+28\cdot\dfrac{1}{(2+1)(2+2)}+8⋅ (1+1)(1+2) 1 ​ +28⋅ (2+1)(2+2) 1 ​ +56\cdot\dfrac{1}{(3+1)(3+2)}+70\cdot\dfrac{1}{(4+1)(4+2)}+56⋅ (3+1)(3+2) 1 ​ +70⋅ (4+1)(4+2) 1 ​ +56\cdot\dfrac{1}{(5+1)(5+2)}+28\cdot\dfrac{1}{(6+1)(6+2)}+56⋅ (5+1)(5+2) 1 ​ +28⋅ (6+1)(6+2) 1 ​ +8\cdot\dfrac{1}{(7+1)(7+2)}+1\cdot\dfrac{1}{(8+1)(8+2)}+8⋅ (7+1)(7+2) 1 ​ +1⋅ (8+1)(8+2) 1 ​ =\dfrac{1013}{90}= 90 1013 ​

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\displaystyle\sum_{j=0}^8(2 𝑗+1 − 2 𝑗 )=\displaystyle\sum_{j=0}^81=9

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Question ID: mtid-5-stid-8-sqid-636-qpid-521