f(x)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.1/(x+9)(x+10) 求f(8)
题目
f(x)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.1/(x+9)(x+10) 求f(8)
答案
f(x)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+.1/(x+9)(x+10) =1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+.+1/(x+9)-1/(x+10)=1/(x+1)-1/(x+10)属于裂项求和∴ f(8)=1/9-1/18=1/18
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