Question

[Trần Khởi My]

1/1+2 + 1/1+2+3 +1/1+2+3+4 + ......+1/1+2+3+4+....+2016

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Answer 1

[soyeon_Tiểu bàng giải]

\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}\)

\(=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{2017}\right)\)

\(=1-\frac{2}{2017}\)

\(=\frac{2015}{2017}\)