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A=111+121+...+701
\(A = \left(\right. \frac{1}{11} + \frac{1}{12} + . . . + \frac{1}{20} \left.\right) + \left(\right. \frac{1}{21} + \frac{1}{22} + . . . + \frac{1}{30} \left.\right)\)
\(+ \left(\right. \frac{1}{31} + \frac{1}{32} + . . . + \frac{1}{40} \left.\right) + \left(\right. \frac{1}{41} + \frac{1}{42} + . . . + \frac{1}{50} \left.\right) + \left(\right. \frac{1}{51} + \frac{1}{52} + . . . + \frac{1}{60} \left.\right)\)
\(+ \left(\right. \frac{1}{61} + \frac{1}{62} + . . . + \frac{1}{70} \left.\right)\)
\(\Rightarrow A < \frac{1}{10} \cdot 10 + \frac{1}{20} \cdot 10 + \frac{1}{30} \cdot 10 + . . . + \frac{1}{60} \cdot 10\)
\(A < 1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{6}\)
\(A < 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \left(\right. \frac{1}{4} + \frac{1}{5} \left.\right)\)
\(A < 2 + 0 , 45 < 2 , 5\)
A= 11 1 + 12 1 +...+ 70 1 A = ( 1 11 + 1 12 + . . . + 1 20 ) + ( 1 21 + 1 22 + . . . + 1 30 ) A=( 11 1 + 12 1 +...+ 20 1 )+( 21 1 + 22 1 +...+ 30 1 ) + ( 1 31 + 1 32 + . . . + 1 40 ) + ( 1 41 + 1 42 + . . . + 1 50 ) + ( 1 51 + 1 52 + . . . + 1 60 ) +( 31 1 + 32 1 +...+ 40 1 )+( 41 1 + 42 1 +...+ 50 1 )+( 51 1 + 52 1 +...+ 60 1 ) + ( 1 61 + 1 62 + . . . + 1 70 ) +( 61 1 + 62 1 +...+ 70 1 ) ⇒ A < 1 10 ⋅ 10 + 1 20 ⋅ 10 + 1 30 ⋅ 10 + . . . + 1 60 ⋅ 10 ⇒A< 10 1 ⋅10+ 20 1 ⋅10+ 30 1 ⋅10+...+ 60 1 ⋅10 A < 1 + 1 2 + 1 3 + . . . + 1 6 A<1+ 2 1 + 3 1 +...+ 6 1 A < 1 + 1 2 + 1 3 + 1 6 + ( 1 4 + 1 5 ) A<1+ 2 1 + 3 1 + 6 1 +( 4 1 + 5 1 ) A < 2 + 0 , 45 < 2 , 5 A<2+0,45<2,5
Đây qu, phiền bạn tick giup mình nha
\(3.M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}\)
=> \(3M-M=2M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{39}}\)
=> \(2M=1-\frac{1}{3^{39}}\)
=> \(M=\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)\)
do \(1-\frac{1}{3^{39}}< 1\)
=> \(\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)< \frac{1}{2}.1=\frac{1}{2}\)
Vay \(M< \frac{1}{2}\)
Chuc bn hoc tot !
a) \(P=\frac{1+2}{1^2.2^2}+\frac{2+3}{2^2.3^2}+...+\frac{9+10}{9^2.10^2}\)
\(P=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\) ( rút gọn số mũ nhé )
\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)
\(P=1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
Vì \(\frac{9}{10}< 1\Rightarrow P< 1\) (đpcm)
b) Chút nữa mình làm nhé ^^
b)
\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
Ta so sánh giữa A và Q.
\(\frac{1}{1.2}>\frac{1}{3};\frac{1}{2.3}>\frac{1}{3^2};\frac{1}{3.4}>\frac{1}{3^3};....;\frac{1}{100.101}>\frac{1}{3^{100}}\)
\(\Rightarrow Q< A\)
Ta lại tiếp tục so sánh A và \(\frac{1}{2}\)
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\Leftrightarrow A< \frac{1}{2}\)
Ta được:
\(Q< A< \frac{1}{2}\Leftrightarrow Q< \frac{1}{2}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< 1-\frac{1}{2017}=\frac{2016}{2017}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{2016}{2017}\left(đpcm\right)\)
Bài làm:
Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\)
=> \(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\right)\)
<=> \(2B=1-\frac{1}{3^{2017}}\)
=> \(B=\frac{1}{2}-\frac{1}{3^{2017}.2}< \frac{1}{2}\)
=> \(B< \frac{1}{2}\)