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分項對消法
Jan 13th 2014, 03:30

設f(x)=1/3√(x+1)平方+3√x(x+1)+3√x平方,則f(1)+f(2)+f(3)+.....+f(999)之值為何

此方法用分項對消來解 雖然有人提出解法,但是完全看不懂

f(x)=1/3√(x+1)平方+3√x(x+1)+3√x平方=(3√x+1-3√x)/〔3√(x+1)平方+3√x(x+1)+3√x平方〕(3√x+1-3√x)=3√x+1-3√x 看不懂這部分

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