Question

[Cute Vô Đối]

1, Tính

a, B=\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + . . . + \(\dfrac{1}{2007.2008}\)

b, Q= \(\dfrac{7}{1.3}\) + \(\dfrac{7}{3.5}\) + . . . + \(\dfrac{7}{2009.2011}\)

c, S= \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ . . . + \(\dfrac{1}{3^{5000}}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)

b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)