\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1

\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)

\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)

\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

= \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\left(đpcm\right)\)

Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}\)

Vì \(\frac{1}{46}>0\Rightarrow1-\frac{1}{46}< 1\)

Vậy \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}< 1\)

Chứng tỏ S < 1

Ủng hộ mk nha ^_^

S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)

  \(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)

   \(=1-\frac{1}{46}=\frac{45}{46}< 1\)

S=3/1.4+3/4.7+3/7.10+.....+3/40.43+3/43.46

S= 1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S= 1-1/46

=> S<1

S=3.(1/1-1/4+1/4-1/7+.........+1/40-1/43+1/43-1/46)          

S=3.(1/1-1/46)

S=3.45/46

S=2/43/46

=> 2/43/46>1

=>S>1

S=3.(1/1-1/4+1/4-1/7+.........+1/40-1/43+1/43-1/46)          

S=3.(1/1-1/46)

S=3.45/46

S=2/43/46

z=2/43/46

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}=\frac{45}{46}\)

biet chet lien luon do 

suy ra k0 biet

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}=1-\frac{1}{46}< 1\)

\(\left(\frac{3}{a\cdot\left(a+3\right)}=\frac{a+3-3}{a\cdot\left(a+3\right)}=\frac{1}{a}-\frac{1}{a+3}\right)\)

\(S=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{43\times46}\)

\(3S=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+...+\frac{3}{43}-\frac{3}{46}\)

\(3S=3-\frac{3}{46}\)

\(3S=\frac{135}{46}\)

\(S=\frac{45}{46}< 1\)

Vậy ra có điều phải chứng minh

= 1 -1/4 +1/4 -1/7 +1/7 -1/10+...+1/40 -1/43 +1/43 -1/46

= 1 -1/46

= 45/46

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}\)

\(=\frac{45}{46}\)

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