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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
Ta có:
S = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}=1-\dfrac{1}{20}< 1\)
Vậy S<1
Ta có:
S = 12+122+123+...+1220<11.2+12.3+13.4+...+119.20=1−120<112+122+123+...+1220<11.2+12.3+13.4+...+119.20=1−120<1
Vậy S<1
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\) . CMR : A<1
Giải:
Có \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\\ \dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\\ ....\\ \dfrac{1}{20^2}< \dfrac{1}{19\cdot20}\)
Nên `A=1/2^2+1/3^2+1/4^2+...+1/(20^2)<1/1.2+1/2.3+1/3.4+...+1/19.20`
`=1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20=1-1/20=19/20`
Mà `19/20<1`
nên `A<1(đpcm)`
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
\(S=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A< 1-\dfrac{1}{50}\Rightarrow A< 1\)
Ta có \(S=\dfrac{1}{2^2}\left(1+A\right)\)
Ta có
\(A< 1\Rightarrow1+A< 2\Rightarrow S< \dfrac{1}{2^2}.2=\dfrac{1}{2}\)
=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022
=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023
=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023
=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022
=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021) - 1/4^2022 - 2023/4^2022 + 2023/4^2023
=> 9S = 4 - 1/4^2022 - 2023/4^2022 + 2023/4^2023
= 4- 2024/4^2022 + 2023/4^2023
Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0
=> 9S < 4 < 9/2
=> S < 1/2 (đpcm)
Ta có S = \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\)
4S = \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\)
4S - S = ( \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\) ) - ( \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\))
3S = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}-\dfrac{2023}{4^{2023}}\)
Đặt A = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\)
4A = 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)
4A - A = ( 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)) - ( 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\))
3A = 4 - \(\dfrac{1}{4^{2022}}\)
A = ( 4 - \(\dfrac{1}{4^{2022}}\)) : 3 = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\)
⇒ 3S = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)
S = ( \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)) : 3 = \(\dfrac{4}{9}-\dfrac{1}{4^{2022}\cdot3^2}-\dfrac{1}{4^{2023}\cdot3}< \dfrac{4}{9}< \dfrac{1}{2}\)
Vậy S < \(\dfrac{1}{2}\)
\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\right)\\ S=1-\dfrac{1}{2^{20}}\\ =>S< 1\\ \)
vậy S bé hơn 1