Cho I =\(\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+.....+\frac{605}{3^{100}}\)
CMR I < 7
K
Khách
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Cho \(I=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+..._{^+}\frac{605}{3^{100}}.\)Chứng minh: \(I<7\)
0
27 tháng 7 2019
a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
27 tháng 7 2019
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
19 tháng 2 2019
Ta có \(H=\frac{7}{3}+\frac{13}{3^2}+...+\frac{605}{3^{100}}\)
\(\Leftrightarrow3H=7+\frac{13}{3}+...+\frac{605}{3^{99}}\)
\(\Rightarrow2H=7+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2H=7+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Mà \(6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)=3-\frac{1}{3^{99}}\)
\(\Rightarrow2H=7+3-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
\(\Leftrightarrow2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2H< 10\)
\(\Leftrightarrow H< 5\left(1\right)\)
Ta có \(2H=10-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Mà\(\frac{1}{3^{97}}+\frac{605}{3^{98}}< 22\)
hay\(\frac{1}{3^{99}}+\frac{605}{3^{98}}< \frac{22}{9}\)
\(\Rightarrow2H>10-\frac{22}{9}=\frac{68}{9}=2\cdot\left(3+\frac{7}{9}\right)\)
\(\Rightarrow H>3+\frac{7}{9}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrowđpcm\)
VM
17 tháng 1 2020
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!
1 tháng 8 2018
a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
17 tháng 12 2016
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
Ta có \(I=\frac{11}{3}+\frac{17}{3^2}+...+\frac{605}{3^{100}}\left(1\right)\)
\(\Leftrightarrow3I=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\left(2\right)\)
Lấy \(\left(2\right)trừ\left(1\right)\)ta có
\(3I-I=11+\frac{6}{3}+\frac{6}{3^2}+...+\frac{6}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=11+6\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{605}{3^{100}}\)
Xét \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\left(3\right)\)
\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{99}}\left(4\right)\)
Lấy\(\left(4\right)-\left(3\right)\)ta có
\(2A=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow6A=3-\frac{1}{3^{99}}\)
Khi đó \(2I=11+3-\frac{1}{3^{99}}-\frac{605}{3^{100}}\)
\(\Leftrightarrow2I=14-\left(\frac{1}{3^{99}}+\frac{605}{3^{100}}\right)\)
Vì\(\frac{1}{3^{99}}+\frac{605}{3^{100}}>0\)
\(\Rightarrow2I< 14\)
\(\Leftrightarrow I< 7\left(đpcm\right)\)