Question

[Phan Cả Phát]

Tính P = \(\dfrac{1^2}{3.5}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}+....+\dfrac{1004^2}{2007.2009}+\dfrac{1005^2}{2009.2011}\)

Answer 1

[Hung nguyen]

\(P=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+...+\dfrac{1005^2}{2009.2011}\)

\(\Leftrightarrow4P=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+...+\dfrac{4.1005^2}{2009.2011}\)

\(=\dfrac{2^2}{2^2-1}+\dfrac{4^2}{4^2-1}+...+\dfrac{2010^2}{2010^2-1}\)

\(=2009+\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2009.2011}\right)\)

\(=2009+\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=2009+\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right)=2009+\dfrac{1005}{2011}\)

Answer 2

[Phan Cả Phát]

Ace Legona Akai Haruma Phương AnPhương AnVõ Đông Anh Tuấn làm jum Hung nguyen