chứng tỏ rằng
a) A= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
b) B= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}< \frac{1}{2}\)
Những câu hỏi liên quan
Chứng tỏ rằng
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
b) \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2018^2}< \frac{3}{4}\)
c) \(\frac{1}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
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Chứng tỏ rằng
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
chứng tỏ rằng :
a) \(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{10}}>\frac{1}{2^{11}}\)
b) \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}>\frac{1}{100}\)
a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm
b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}
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Câu 1: Tính: Afrac{1+left(1+2right)+left(1+2+3right)+...+left(1+2+3+...+2017right)}{1cdot2+2cdot3+3cdot4+...+2017cdot2018}Câu 2: Cho: Afrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8} và Bfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}Câu 3: Chứng tỏ rằng: frac{1}{3}+frac{1}{31}+frac{1}{35}+frac{1}{37}+frac{1}{47}+frac{1}{53}+frac{1}{61} frac{1}{2}Câu 4: Tìm các số tự nhiên a, b sao cho: frac{a}{2}+frac{b}{3}frac{a+b}{2+3}Câu 5: Tính Aleft(frac{1}{2^2}-1right)cdotleft(frac{1}{3^2}-1right)cdotleft(frac{1}{4^2}-1rig...
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Câu 1: Tính: \(A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{1\cdot2+2\cdot3+3\cdot4+...+2017\cdot2018}\)
Câu 2: Cho: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\) và \(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
Câu 3: Chứng tỏ rằng: \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)
Câu 4: Tìm các số tự nhiên a, b sao cho: \(\frac{a}{2}+\frac{b}{3}=\frac{a+b}{2+3}\)
Câu 5: Tính \(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{100^2}-1\right)\)
Câu 6: Tìm số tự nhiên n để các phân số tối giản
\(A=\frac{2n+3}{3n-1}\), \(B=\frac{3n+2}{7n+1}\)
Câu 7: So sánh: \(A=1\cdot3\cdot5\cdot7\cdot...\cdot99\) với \(B=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}\)
Câu 8: Chứng tỏ rằng:
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}< 1\)
b) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
Câu 9: Cho \(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\)
Chứng minh rằng: \(\frac{1}{3}< A< \frac{1}{2}\)
Câu 10: Chứng tỏ rằng: \(\frac{7}{12}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}< 1\)
Câu 8( Mình không viết đè nữa nha)
a) 2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100
= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100
= 1 – 1/100 < 1
= 99/100 < 1
Vậy A< 1
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Cho 3 số a;b;c thỏa mãn a.b.c1Chứng minhfrac{1}{ab+a+1}+frac{b}{bc+b+1}+frac{1}{abc+bc+b}12)CHứng tỏ rằnga1-frac{1}{2}-frac{1}{2^2}-frac{1}{2^3}-....-frac{1}{2^{10}}frac{1}{2^{11}}b1-frac{1}{2^2}-frac{1}{3^2}-frac{1}{4^2}-...-frac{1}{100^2}frac{1}{100}cleft(frac{1}{2^2}-1right).left(frac{1}{3^2}-1right).left(frac{1}{4^2}-1right)...left(frac{1}{100^2}-1right)
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Cho 3 số a;b;c thỏa mãn a.b.c=1
Chứng minh
\(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}=1\)
2)
CHứng tỏ rằng
\(a=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}>\frac{1}{2^{11}}\)
\(b=1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}>\frac{1}{100}\)
\(c=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)
1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)
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chứng tỏ rằng
C = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
D = \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Phần C đề thiếu
\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D=3-\frac{203}{3^{100}}\)
\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)
\(C=\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(\Rightarrow2C=1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(\Rightarrow2C+C=(1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}})+\)\((\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}})\)
\(\Rightarrow3C=1-\frac{1}{100}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{300}< \frac{1}{3}\left(đpcm\right)\)
Chứng tỏ rằng: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Làm theo cách của Trắng nha ,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)\)
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Ta có: \(\frac{1}{2^2}=\frac{1}{2^2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...................
\(\frac{1}{2019^2}< \frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}\)
\(=\frac{3}{4}-\frac{1}{2019}\)\(< \frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Điều phải chứng minh
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Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}\)
Ta có:
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
....
\(\frac{1}{2019^2}=\frac{1}{2019.2019}< \frac{1}{2018.2019}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow A< 1-\frac{1}{2019}\)
\(\Rightarrow A< \frac{2018}{2019}\)
đến đây mới thấy mik sai ,xin lỗi
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Chứng tỏ rằng:a/ frac{1}{2} frac{1}{41}+frac{1}{42}+frac{1}{43}+...+frac{1}{80} 1b/ 1 frac{3}{10}+frac{3}{11}+frac{3}{12}+frac{3}{13}+frac{3}{14} 2c/ Afrac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+...+frac{1}{2^{100}} 1d/ Bfrac{1}{3}+frac{1}{3^2}+frac{1}{3^3}+...+frac{1}{3^{99}} frac{1}{2}e/ frac{2}{5} frac{1}{41}+frac{1}{42}+frac{1}{43}+...+frac{1}{80} frac{2}{3}f/Cfrac{3}{1^2cdot2^2}+frac{5}{2^2cdot3^2}+frac{7}{3^2cdot4^2}+...+frac{19}{9^2cdot10^2} 1
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Chứng tỏ rằng:
a/ \(\frac{1}{2}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}< 1\)
b/ \(1< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\)
c/ A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}< 1\)
d/ \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}< \frac{1}{2}\)
e/ \(\frac{2}{5}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}< \frac{2}{3}\)
f/\(C=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{19}{9^2\cdot10^2}< 1\)
\(b)\) Đặt \(B=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) ta có :
\(B>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{3+3+3+3+3}{15}=\frac{3.5}{15}=\frac{15}{15}=1\)
\(\Rightarrow\)\(B>1\) \(\left(1\right)\)
Lại có :
\(B< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3+3+3+3+3}{10}=\frac{3.5}{10}=\frac{15}{10}< \frac{20}{10}=2\)
\(\Rightarrow\)\(B< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(1< B< 2\) ( đpcm )
Vậy \(1< B< 2\)
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\(a)\) Đặt \(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\) ta có :
\(A>\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\)
Do từ \(41\) đến \(80\) có \(\left(80-41\right):1+1=40\) số nên có \(40\) phân số \(\frac{1}{80}\) suy ra :
\(A>40.\frac{1}{80}=\frac{40}{80}=\frac{1}{2}\)
\(\Rightarrow\)\(A>\frac{1}{2}\) \(\left(1\right)\)
Lại có :
\(A< \frac{1}{41}+\frac{1}{41}+\frac{1}{41}+...+\frac{1}{41}\)
Do từ \(41\) đến \(80\) có \(\left(80-41\right):1+1=40\) số nên có \(40\) phân số \(\frac{1}{41}\) suy ra :
\(A< 40.\frac{1}{41}=\frac{40}{41}< 1\)
\(\Rightarrow\)\(A< 1\) \(\left(2\right)\)
Từ (1) và (2) suy ra :
\(\frac{1}{2}< A< 1\) ( đpcm )
Vậy \(\frac{1}{2}< A< 1\)
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Chứng tỏ rằng;
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2019^2}\)
\(\Rightarrow A=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\)
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đặt A=1/2^2+....+1/2019^2
vì 1/2^2+....+1/2019^2<1/1.2+1/2.3+....+1/2018.2019
=> A<1/1-1/2+1/2-1/3+.....+1/2018-1/2019
=> A<1-1/2019=2018/2019<3/4.
=> A<3/4.
vậy 1/2^2+....+1/2019^2<3/4
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Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)\(+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{2018}{2019}\)
Mà: \(\frac{3}{4}=\frac{2016}{2688}< \frac{2017}{2688}< \frac{2017}{2019}< \frac{2018}{2019}\)
\(\Rightarrow\frac{3}{4}< \frac{2018}{2019}\)
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