2. Chứng minh
a, \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{50^2}\) < 1
b, \(\dfrac{1}{3}\)< \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{150}\)< \(\dfrac{1}{2}\)
A\(=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\). Chứng minh A < 2
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{50^2-1}\)
\(\Leftrightarrow A< 1+\dfrac{1}{3}+\dfrac{1}{8}+...+\dfrac{1}{2499}\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+...++\dfrac{1}{48}-\dfrac{1}{50}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(1-\dfrac{1}{51}+\dfrac{1}{2}-\dfrac{1}{50}\right)\)
\(\Leftrightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)\)
Nhận xét \(\dfrac{50}{51}< 1;\dfrac{24}{50}< 1\Rightarrow A< 1+\dfrac{1}{2}\cdot\left(\dfrac{50}{51}+\dfrac{24}{50}\right)< 1+\dfrac{1}{2}\cdot\left(1+1\right)=2\)
Vậy A<2
Nhận xét: \(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...........
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
Vậy A < 2
Ta có:
\(\dfrac{1}{1^2}=1;\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{40.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\)
Cho A =\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\) .Chứng minh A < 2
Ta có :
\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+......................+\dfrac{1}{50^2}\)
Ta thấy :
\(\dfrac{1}{1^2}=1\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
............................
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....................+\dfrac{1}{49.50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< 1+1-\dfrac{1}{50}\)
\(\Rightarrow A< 2-\dfrac{1}{50}< 2\)
\(\Rightarrow A< 2\rightarrowđpcm\)
S = \(\dfrac{1}{1^2}\) +\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+....+\(\dfrac{1}{50^2}\). Chứng minh rằng S < 2
1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50
=1/1-1/2+1/2-1/3+...+1/49-1/50<1
=>S<1+1=2
Cho A=1+\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2^{100}-1}\)
Chứng minh rằng 50<A<100
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
a , cho A = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\) . Chứng minh A < \(\dfrac{7}{4}\)
b ,cho B = 21 + 22 + 23 + ... + 260 . Chứng minh B \(⋮\) 21
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
Cho: A= 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2^{100}-1}\)
Chứng minh rằng: 50 < A < 100
Giúp mình với!
8 Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) ; b) \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100}{100!}\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
d) \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}< \dfrac{1}{2}\)
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
Đặt A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.......+\dfrac{1}{3^{99}}\)
=> 3A=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+..........+\dfrac{1}{3^{98}}\)
=> 3A-A= 1-\(\dfrac{1}{3^{99}}\)
=> A=\(\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)
=> A<1/2
Vậy A<1/2
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)