Question

bieu thuc A =\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+...+\(\frac{1}{1+2+3+4...+2014}\)co gia tri bang 

2lk may bn oi nhanh gium mk vs 

Answer 1

\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\)

\(\Rightarrow2A=2\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2029105}\right)\)

\(\Rightarrow2A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{4058210}\)

\(\Rightarrow2A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2015}\)

\(\Rightarrow2A=\frac{2013}{4030}\)

\(\Rightarrow A=\frac{2013}{8060}\)

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