cả 2 cái cộng lại hay là từng cái một vậy bạn?

a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?

\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)

b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)

=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)

=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)

\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)

\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)

\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)  

\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)

\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)

\(S=\frac{1}{2}-\frac{1}{32}\)

\(S=\frac{17}{32}< 1\)

S=6/2.5+6/5.8+6/8.11+...+6/29.32
S=2.(1/2-1/5+1/5-1/8+1/8-1/11+...+1/29-1/32)
S=2.(1/2+1/32)
S=2.15/32
S=30/32=15/16
=>S<1

\(S=2.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=1-\frac{1}{16}< 1\)

Vậy \(S< 1\)

Bài 1 :

S = \(\frac{6}{2.5}+\frac{6}{5.8}+...+\frac{6}{29.32}\)

   = 2 . \(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)

   = 2 . \(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

   = 2 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....

S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)

   = 1/3 . (1/2 - 1/20)

   = 1/3 . 9/20

   = 3/20

\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)

\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(S=\frac{9}{20}:3=\frac{3}{20}\)

\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...........+\frac{3}{17.20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{10}{20}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\frac{9}{20}\)

\(-\frac{3}{20}\)

3S=3/2.5+3/5.8+3/8.11+...+3/101.104

3S=1/2-1/5+1/5-1/8+1/8-1/11+...+1/101-1/104

3S=1/2-1/104

S=51/104:3

S=17/104

Vậy S=17/104

         \(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{101.104}\)

\(\Rightarrow3S=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.......+\frac{1}{101.104}\right)\)

           \(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{101.104}\)

           \(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{101}-\frac{1}{104}\)

           \(=\frac{1}{2}-\frac{1}{104}\)

           \(=\frac{51}{104}\)

           \(\Rightarrow S=\frac{51}{104}:3=\frac{51}{104}.\frac{1}{3}\)

                     \(=\frac{7}{104}\)

                VẬY   \(S=\frac{7}{104}\)

Ta có: 3S = 3/2.5 + 3/5.8 + ... + 3/47.50

           3S = 1/2 - 1/5 + 1/5 - 1/8 + ... +1/47 - 1/50

           3S = 1/2 - 1/50

           3S = 12/25

           => S = 12/25 : 3 = 4/25 

k, đây là dạng toán sai phân hữu hạn. 
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số hạng tổng quát là 1/[n.(n+3)] = (1/3).[(n+3)-n]/[n.(n+3)] = (1/3). [1/n - 1/(n+3)] 
=> 
A = (1/3).[(1/2 - 1/5) + (1/5 - 1/8) + (1/8 - 1/11) +...+(1/44 - 1/47) + (1/47 - 1/50)] 
= (1/3).[1/2 - 1/50] 
= (1/3). (24/50) = (1/3).(12/25) = 4/25 
vậy A = 4/25 
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good luck!

Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)

\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)

Vậy:..........................................(đpcm)

a/ Ta có: \(S=1+\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{27}-\frac{1}{30}\right)\)

\(S=1+\left(\frac{1}{2}-\frac{1}{30}\right)\)

\(S=1+\frac{7}{15}\)

\(S=\frac{22}{15}\)

b/ \(S=-4+\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{107}-\frac{1}{110}\right)\)

\(S=-4+\left(1-\frac{1}{110}\right)\)

\(S=-4+\frac{109}{110}\)

\(S=-3\frac{1}{110}\)