Cho \(S=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\). Chứng tỏ rằng S<\(\dfrac{1}{16}\)
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23 tháng 4 2023 lúc 15:20
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\) . Chứng tỏ rằng \(S< \dfrac{1}{16}\)
9 tháng 5 2022 lúc 20:42
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
Chứng tỏ rằng :\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
Ta có :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{2}{5}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.................+\dfrac{99}{100}}\)
\(=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+.............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+............+1-\dfrac{1}{100}}\)
\(=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+...........+\dfrac{2}{100}\right)}{\left(1+1+.........+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{100}\right)}\)
\(=\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..........+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.........+\dfrac{1}{100}\right)}\)
\(=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+..........+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}=2\rightarrowđpcm\)
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Chứng tỏ rằng: \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
11 tháng 4 2021 lúc 15:06
text{Bài 4. Chứng tỏ rằng:}a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{30^2} 1b) dfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1c) dfrac{1}{5}+dfrac{1}{6}+dfrac{1}{7}+...+dfrac{1}{17} 2d) dfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{29.30} 1
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\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
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c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
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Câu 1 : Thực hiện phép tính 1 cách hợp lý :
a) dfrac{-12}{7}.dfrac{4}{35}+dfrac{12}{7}.dfrac{left(-31right)}{35}-dfrac{2}{7}
b) 1+2-3-4+5+5-7-8+...+97+98-99-100
c) A157.left(-37right)-left(41.53-37.157right)+51.53
d) Bleft(dfrac{1}{11}+dfrac{1}{21}+dfrac{1}{31}+dfrac{1}{41}+dfrac{1}{51}right)left(dfrac{-41}{123}+dfrac{31}{-186}-dfrac{-51}{102}right)
Câu 2 :
a) 12 ( x - 5 ) 7x - 5
b) Tìm x in Z sao cho : ( 2x - 3 ) 2010 ( 2x - 3 ) 2012
Câu 3 :
1) Cho biểu thức S 1 + 3 + 32 + 33 +.....
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Câu 1 : Thực hiện phép tính 1 cách hợp lý :
a) \(\dfrac{-12}{7}.\dfrac{4}{35}+\dfrac{12}{7}.\dfrac{\left(-31\right)}{35}-\dfrac{2}{7}\)
b) \(1+2-3-4+5+5-7-8+...+97+98-99-100\)
c) \(A=157.\left(-37\right)-\left(41.53-37.157\right)+51.53\)
d) \(B=\left(\dfrac{1}{11}+\dfrac{1}{21}+\dfrac{1}{31}+\dfrac{1}{41}+\dfrac{1}{51}\right)\left(\dfrac{-41}{123}+\dfrac{31}{-186}-\dfrac{-51}{102}\right)\)
Câu 2 :
a) 12 ( x - 5 ) = 7x - 5
b) Tìm x \(\in\) Z sao cho : ( 2x - 3 ) 2010 = ( 2x - 3 ) 2012
Câu 3 :
1) Cho biểu thức S = 1 + 3 + 32 + 33 +...+ 3202 + 3 203
a) chứng tỏ rằng tổng S chia hết cho 52 .
b) Tìm Chữ số tận cùng trong tổng S .
2 ) Cho biểu thức A= \(\dfrac{2n+1}{2n+5}\) . Chứng tỏ rằng với mọi số tự nhiên n thì A là phân số tối giản .
Câu 4 : So sánh tổng gồm 1006 số hạng :
\(S=\dfrac{1}{1.1.3}+\dfrac{1}{2.3.5}+\dfrac{1}{3.5.7}+...+\dfrac{1}{1006.2011.2013}\) với \(\dfrac{2}{3}\)
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
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Chứng minh rằng :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\)
Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
1) Tính hợp lí
a) 5 + 53 + 55 + ... + 597 + 599
2) Cho A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Chứng tỏ rằng \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
2. Chứng tỏ:\(\dfrac{2}{5}< A< \dfrac{8}{9}.\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)
Giải:
Ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}.\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}.\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}.\)
\(A< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}.\)
\(A< 1+0+0+0+...+0-\dfrac{1}{9}.\)
\(A< 1-\dfrac{1}{9}.\)
\(A< \dfrac{8}{9}_{\left(1\right)}.\)
Ta lại có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}.\)
\(A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}.\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}.\)
\(A>\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-\dfrac{1}{10}.\)
\(A>\dfrac{1}{2}+0+0+0+...+\dfrac{1}{10}.\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}.\)
\(A>\dfrac{4}{10}.\)
\(\Rightarrow A>\dfrac{2}{5}_{\left(2\right)}.\) (vì \(\dfrac{4}{10}=\dfrac{2}{5}.\))
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\).
\(\Rightarrow A< \dfrac{8}{9}\) và \(A>\dfrac{2}{5}.\)
\(\Rightarrow\) \(\dfrac{8}{9}>A>\dfrac{2}{5}\) hay \(\dfrac{2}{5}< A< \dfrac{8}{9}.\)
Vậy ta thu được \(đpcm.\)
~ Học tốt!!!... ~ ^ _ ^
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Sr bn vì bây giờ mik ms nghĩ ra phần a, hơi lâu, chẳng bt bn có cần ns ko, nhưng mik cứ lm giúp bn z!!! *buồn*
Giải:
a, \(A=5+5^3+5^5+...+5^{97}+5^{99}.\)
\(25A=25\left(5+5^3+5^5+...+5^{97}+5^{99}\right).\)
\(25A=5^2\left(5+5^3+5^5+...+5^{97}+5^{99}\right).\)
\(25A=5^3+5^5+5^7+...+5^{99}+5^{101}.\)
\(25A-A=\left(5^3+5^5+5^7+...+5^{99}+5^{101}\right)-\left(5+5^3+5^5+...+5^{97}+5^{99}\right).\)
\(24A=\left(5^{101}-5\right)+\left(5^3-5^3\right)+\left(5^5-5^5\right)+...+\left(5^{97}-5^{97}\right)+\left(5^{99}-5^{99}\right).\)
\(24A=\left(5^{101}-5\right)+0+0+...+0+0.\)
\(24A=5^{101}-5.\)
\(\Rightarrow A=\dfrac{5^{101}-5}{24}.\)
Vậy \(A=\dfrac{5^{101}-5}{24}.\)
~ Học tốt!!! ~ ^ _ ^
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