Question

[The Smosh]

A=1/1.2+1/3.4+...+1/2005.2006 và B=1/1004.2006+1/1005.2006+...+1/2006.1004

CMR A/B thuộc Z

Answer 1

[Phùng Minh Quân]

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005.2006}\)

\(A=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)

\(A=\left(1+\frac{1}{2}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1003}\right)\)

\(A=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)

\(B=\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.1004}\)

\(3010B=\frac{1004+2006}{1004.2006}+\frac{1005+2005}{1005.2005}+...+\frac{2006+1004}{2006.1004}\) ( sửa đề nhé ) 

\(3010B=\frac{1}{2006}+\frac{1}{1004}+\frac{1}{2005}+\frac{1}{1005}+...+\frac{1}{1004}+\frac{1}{2006}\)

\(3010B=2\left(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\right)\)

\(B=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}\)

\(\Rightarrow\)\(\frac{A}{B}=\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{\frac{\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}}{1505}}=1505\) hay \(\frac{A}{B}\inℤ\)

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