Question

[Tiểu Thư Họ Đỗ]

1. Tính

a. S = \(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+.......+\(\dfrac{1}{2002.2005}\)

b. P = \(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+.......+\(\dfrac{3}{96.101}\)

c. Q = \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+.......+\(\dfrac{1}{98.99.100}\)

Help me!

Answer 1

[Lucy Heartfilia]

S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)

S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)

S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)

S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)

S = \(\dfrac{2014}{6015}\)

Answer 2

[Hoàng Thị Ngọc Anh]

a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)

KL.

b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)

\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)

KL.

c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)

\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)

KL.