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2T
14 tháng 8 2019
\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)
\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)
Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
27 tháng 12 2016
\(\frac{1}{2^2}< \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)\)
\(\frac{1}{3^2}< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(\frac{1}{4^2}< \frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(\frac{1}{5^2}< \frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)\)
....
\(\frac{1}{1990^2}< \frac{1}{2}\left(\frac{1}{1989}-\frac{1}{1991}\right)\)
công hết lại: ra điều cần chứng minh
cho @ ...thêm cái nữa
\(\frac{1}{n^2}< \frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n-2}\right)\)
10 tháng 1 2023
a: \(\dfrac{3}{4}A=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+...+\left(\dfrac{3}{4}\right)^{2021}\)
=>\(\dfrac{7}{4}\cdot A=\left(\dfrac{3}{4}\right)^{2021}+1\)
=>\(A\cdot\dfrac{7}{4}=\dfrac{3^{2021}+4^{2021}}{4^{2021}}\)
=>\(A=\dfrac{3^{2021}+4^{2021}}{4^{2020}\cdot7}\)
b: Vì 3^2021+4^2021 ko chia hết cho 4^2020*7 nên A ko là số nguyên
S
3 tháng 5 2019
S = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/2019^2
S = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/2019^2)
S = 2018 - (1/4 + 1/9 + 1/16 + ... + 1/2019^2)
đặt A = 1/4 + 1/9 + 1/16 + ... + 1/2019^2
có : 1/4 = 1/2*2 < 1/1*2
1/9 = 1/3*3 < 1/2*3
...
1/2019^2 < 1/2018*2019
=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + /12018*2019
=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4+ ... + 1/2018 - 1/2019
=> A < 1 - 1/2019
=> A < 2018/2019
=> A không phải số nguyên
S = 2018 - A
=> S không phải 1 số nguyên
PT
1
BT
0
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2019
\(S=\dfrac{1}{2018}\left(1+\dfrac{1}{1}+1+\dfrac{1}{2}+1+\dfrac{1}{3}+...+1+\dfrac{1}{2018}\right)\)
\(S=\dfrac{1}{2018}\left(2018+\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
\(S=1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
Do \(\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2018}\right)>0\Rightarrow S>1\) (1)
Lại có:
\(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}< \dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+...+\dfrac{1}{1}=2018\)
\(\Rightarrow1+\dfrac{1}{2018}\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)< 1+\dfrac{1}{2018}.2018=2\)
\(\Rightarrow S< 2\) (2)
Từ (1), (2) \(\Rightarrow1< S< 2\)
\(\Rightarrow S\) nằm giữa 2 số tự nhiên liên tiếp nên S không phải là số tự nhiên
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2019}{4^{2019}}\)
\(\Rightarrow4S=4(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2019}{4^{2019}})\)
\(\Rightarrow4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}\)
\(\Rightarrow4S-S=(1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}})-(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2019}{4^{2019}})\)
\(\Rightarrow3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow3S< 1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\left(1\right)\)
Đặt: \(A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4A=4(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}})\)
\(\Rightarrow4A=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4A-A=(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2017}})-(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}})\)
\(\Rightarrow3A=4+\frac{1}{4^{2018}}\)
\(\Rightarrow A=\frac{4}{3}+\frac{1}{4^{2018}.3}\)
\(\Rightarrow A< \frac{4}{3}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow3S< \frac{4}{3}\)
\(\Rightarrow S< \frac{4}{9}\Rightarrow S< \frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S< \frac{1}{2}\left(đpcm\right)\)