ND

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2017)(x+2018)

 

KB
19 tháng 12 2017 lúc 16:19

Tính tổng

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2017)(x+2018)

Giải:\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+2017\right)\left(x+2018\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+..........+\frac{1}{x+2017}-\frac{1}{x+2018}\)

\(=\frac{1}{x}-\frac{1}{x+2018}\)

Vậy........................................

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