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\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}=1-\frac{1}{46}< 1\)
\(\left(\frac{3}{a\cdot\left(a+3\right)}=\frac{a+3-3}{a\cdot\left(a+3\right)}=\frac{1}{a}-\frac{1}{a+3}\right)\)
\(S=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{43\times46}\)
\(3S=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+...+\frac{3}{43}-\frac{3}{46}\)
\(3S=3-\frac{3}{46}\)
\(3S=\frac{135}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy ra có điều phải chứng minh
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1
\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)
\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)
\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
= \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(s=1-\frac{1}{46}< 1\)
Vậy S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{43\cdot46}\)
\(S=1\left[\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{43\cdot46}\right]\)
\(S=1\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\right]\)
\(S=1\left[1-\frac{1}{46}\right]=1\cdot\frac{45}{46}=\frac{45}{46}< 1(đpcm)\)
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 \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S< 1\)
Chúc bạn học tốt !!!
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
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}=\frac{45}{46}\)
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