Tính :
a. F = \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+....+\dfrac{4}{2008.2010}\)
b. C = \(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+....+\dfrac{1}{990}\)
c. T = \(1+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+....+\dfrac{1}{16}.\left(1+2+3+....+16\right)\)
d. H = \(\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
( Các bạn giúp mk , làm được 1 câu cũng được , làm hết thì càng tốt )
\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
b, C=\(\dfrac{1}{18}\)+\(\dfrac{1}{54}\)+....+\(\dfrac{1}{990}\)
\(\Rightarrow\)C=\(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\)
=>3C=3( \(\dfrac{1}{3.6}\)+\(\dfrac{1}{6.9}\)+...+\(\dfrac{1}{30.33}\))
=>3C=\(\dfrac{3}{3.6}\)+\(\dfrac{3}{6.9}\)+....+\(\dfrac{3}{30.33}\)
=> 3C=\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{9}\)+...+\(\dfrac{1}{30}\)-\(\dfrac{1}{33}\)
=> 3C= \(\dfrac{1}{3}\)-\(\dfrac{1}{33}\)
=>3C=\(\dfrac{10}{33}\)
=> C=\(\dfrac{10}{33}\):3
=> C=\(\dfrac{10}{99}\)
a, F=4/2.4+4/4/6+4/6.8+......+4/2008.2010
=> F= 4/2.(2/2.4+2/4.6+2/6.8+......+2/2008/2010
=> F= 4/2. ( 1/2-1/4+1/4-1/6+1/6-1/8+......+1/2008-1/2010
=> F=4/2.( 1/2-1/2010)
=> F= 4/2. 502/1005
=> F= \(\dfrac{1004}{1005}\)
