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!
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 = \(\dfrac{2014}{6015}\)
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.
\(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+......+\dfrac{1}{98.99.100}\)
\(2Q=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+........+\dfrac{2}{98.99.100}\)
\(2Q=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
(do \(\dfrac{2}{a.\left(a+1\right).\left(a+2\right)}=\dfrac{1}{a.\left(a+1\right)}-\dfrac{1}{\left(a+1\right).\left(a+2\right)}\))
\(2Q=\dfrac{1}{1.2}-\dfrac{1}{99.100}\)
\(2Q=\dfrac{1}{2}-\dfrac{1}{9900}=\dfrac{4949}{9900}\)
\(Q=\dfrac{4949}{9900}:2=\dfrac{4949}{19800}\)
Chúc bạn học tốt!!!
b , P = \(\dfrac{3}{1.6}\)+ \(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+ ... + \(\dfrac{3}{96.101}\)
P = ( \(\dfrac{1}{1.6}\)+\(\dfrac{1}{6.11}\)+\(\dfrac{1}{11.16}\)+...+\(\dfrac{1}{96.101}\)) . \(\dfrac{3}{5}\)
P = (1 - \(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\)- \(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)- ... + \(\dfrac{1}{96}\)- \(\dfrac{1}{101}\)) . \(\dfrac{3}{5}\)
P = ( 1 - \(\dfrac{1}{101}\)) . \(\dfrac{3}{5}\)
P = \(\dfrac{100}{101}\) . \(\dfrac{3}{5}\)
P = \(\dfrac{100.3}{101.5}\)= \(\dfrac{20.3}{101.1}\)= \(\dfrac{60}{505}\)
\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2002.2005}\)
\(S=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+...+\dfrac{2005-2002}{2002.2005}\)
\(3S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\)
\(3S=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(\Rightarrow S=\dfrac{2004}{2005}\div3=\dfrac{668}{2005}\)
\(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(P=\dfrac{3}{5}.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(P=\dfrac{3}{5}.\left(1-\dfrac{1}{101}\right)\)
\(P=\dfrac{3}{5}\cdot\dfrac{100}{101}=\dfrac{60}{101}\)
\(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(2Q=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
\(2Q=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)
\(\Rightarrow Q=\dfrac{4949}{9900}\div2=\dfrac{4949}{19800}\)
