Chứng minh rằng: 1/22 + 1/32 + 1/42 + ... + 1/1002 < 1
Nhanh dùm nha các bạn. 
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NH
0
VM
22 tháng 4 2016
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 > 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 > 1/3- 1/4+1/4-1/5+1/501/6+1/6-1/7+1/7-1/8+1/8-1/9
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 > 3/9-1/9
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 > 2/9
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 < 1/2.3 +1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 < 1/2 -1/3+1/3-1/4+1/4-1/5+1/6-1/7+1/7-1/8
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 < 1/2-1/8
1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 < 3/8 vậy ta có 2/9< 1/3^2+1/4^2+1/5^2+1/6^2 +1/7^2 1/8^2 <3/8
22 tháng 4 2016
thank u bạn Vũ Minh DŨng nhìu nha mình tịk cho pạn zồi ak nha
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \)\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\left(1\right)\)
Mà \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(2\right)\). Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(A< B< 1\Rightarrow A< 1\)