Tính tổng: S=2/3.7+2/7.1+2/11.15+...+2/95.99
M=1/2+1/6+1/12+1/20+1/110
K
Khách
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16 tháng 7 2017
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
LT
Tính tổng:a)3+3/5+3/25+3/125+3/625
b)M=4/3.7+4/7.11+4/11.15+...+8/95.99
c)N=1/2+1/6+1/12+1/20+...+1/90
1
14 tháng 7 2017
Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
23 tháng 2 2020
Ta có:
\(C=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{40.41}+\frac{2}{41.42}\)
\(\Rightarrow C=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}+\frac{1}{41.42}\right)\)
\(\Rightarrow C=2\left(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{41-40}{40.41}+\frac{42-41}{41.42}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}+\frac{1}{41}-\frac{1}{42}\right)\)
\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{42}\right)=\frac{13}{21}\)
\(D=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(\Rightarrow D=\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{111-107}{107.111}\)
\(\Rightarrow D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)\(E=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow E=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}+\frac{11-10}{10.11}\)
\(\Rightarrow E=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
10 tháng 11 2019
mk làm phần a thui nhé
a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
A = 1/2 - 1/6
A= 3/6 - 1/6
A = 1/3
10 tháng 11 2019
\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)
\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(b=\frac{1}{2}-\frac{1}{14}\)
\(b=\frac{3}{7}\)
\(d=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(d=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(d=1-\frac{1}{11}\)
\(d=\frac{10}{11}\)
\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)
SL
2
8 tháng 5 2021
Giải:
A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)
A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)
A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)
A=1/2.(1/3-1/n+4)
A=1/6-1/2.(n+4)
⇒A<1/6
Chúc bạn học tốt!
8 tháng 5 2021
Ta có : \(A=\dfrac{2}{3.7}+\dfrac{2}{7.11}+...+\dfrac{2}{n\left(n+4\right)}\)
\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)
\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)\(=\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{2}{11}+...+\dfrac{2}{n}-\dfrac{2}{n+4}=\dfrac{2}{3}-\dfrac{2}{n+4}\)
\(\Rightarrow A=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}\)
- Xét hiệu \(A-\dfrac{1}{6}=-\dfrac{1}{2\left(n+4\right)}< 0\)
Vậy A < 1/6
TB
1
Y
24 tháng 6 2019
Chắc là đề thiếu: \(y=\frac{1}{2}-\frac{1}{3\cdot7}-\frac{1}{7\cdot11}-\frac{1}{11\cdot15}-\frac{1}{15\cdot19}-\frac{1}{19\cdot23}-\frac{1}{23\cdot27}\)
\(y=\frac{1}{2}-\left(\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+...+\frac{1}{23\cdot27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{23\cdot27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}\right)\)
\(=\frac{1}{2}-\frac{1}{4}\left(\frac{1}{3}-\frac{1}{27}\right)=\frac{1}{2}-\frac{1}{4}\cdot\frac{8}{27}=\frac{23}{54}\)
AM
Tính tổng
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{110}+\frac{1}{132}\)
7
2 tháng 4 2017
=1/1*2+1/2*3+1/3*4+...+1*10*11+1/11*12=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11+1/11-1/12
=1-1/12=11/12.
DP
2 tháng 4 2017
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{10\times11}+\frac{1}{11\times12}\)
\(=1-\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{11}+\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
S=2/3.7+2/7.1+2/11.15+...+2/95.99
S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{2}.\left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{15}\right)+...+\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{99}\right)\)
S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)
S = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
S = \(\frac{1}{2}.\frac{32}{99}\)
S = \(\frac{16}{99}\)
M=1/2+1/6+1/12+1/20+..+1/110
M = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
M = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
M = \(1-\frac{1}{11}\)
M = \(\frac{10}{11}\)