chứng tỏ tổng sau nhỏ hơn 1:A=3/1x4+3/4x7+3/7x10+.....+3/40x43
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ND
5
23 tháng 4 2016
Cho S=3/1x4+3/4x7+3/7x10+...+3/40x43+3/43x46. Hãy chứng tỏ S<1
ĐPM : S < 1
TN
23 tháng 4 2016
S=3/1x4+3/4x7+3/7x10+...+3/40x43+3/43x46
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
=>S<1
DH
1
1 tháng 5 2015
= 1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
= 1 - 1/46 = 45/46 < 1
NM
3
21 tháng 1 2020
Ta có:
S=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
S=\(1-\frac{1}{n+3}\)
=>S<1
Vậy S<1
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
VH
4
LP
2
XO
6 tháng 6 2021
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}=1-\frac{1}{46}=\frac{45}{46}\)
QA
6 tháng 6 2021
Trả lời:
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}+\frac{3}{43\cdot46}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=\frac{1}{1}-\frac{1}{46}\)
\(=\frac{46}{46}-\frac{1}{46}\)
\(=\frac{45}{46}\)
ML
3 tháng 8 2017
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(\Leftrightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(\Leftrightarrow A=1-\frac{1}{46}\)
\(\Leftrightarrow A=\frac{45}{46}\)
VA
21 tháng 7 2016
1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
21 tháng 7 2016
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
9 tháng 5 2016
\(A=3\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{97\times100}\right)\)
\(A=3\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3\times\left(1-\frac{1}{100}\right)\)
\(A=3\times\frac{99}{100}\)
\(A=\frac{297}{100}\)
HP
9 tháng 5 2016
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+......+\frac{3^2}{97.100}\)
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Đặt \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
Ta có: \(S=\frac{3}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{97.100}\right)\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{97}-\frac{1}{100}\)
\(S=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=3.S=3.\frac{99}{100}=\frac{297}{100}\)
\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{40.43}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{40}-\dfrac{1}{43}\)
\(A=1-\dfrac{1}{43}\)
\(A< 1\left(đpcm\right)\)
\(A=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)\)
\(=3\left(1-\dfrac{1}{43}\right)=\dfrac{126}{43}>1\)
... sai đâu không nhỉ??