chứng minh S = 1/4 +2/4^2 +3/4^3 + ... + 2019/4^21019 nhỏ hơn 1/2
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TC
2
21 tháng 3 2019
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}.\)
\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}-\frac{1}{4}-\frac{2}{4^2}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)
\(3S< A=1+\frac{1}{4}+...+\frac{1}{4^{2018}}\)\(\Rightarrow3A=4A-A=4-\frac{1}{4^{2018}}< 4\)(sau khi rút gọn)
\(\Rightarrow3.3S< 4\Rightarrow9S< 4\)
\(\Rightarrow S< \frac{4}{9}< \frac{1}{2}\)
TN
1
NH
1
AH
Akai Haruma
Giáo viên
31 tháng 3 2023
Lời giải:
$A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1000^2}$
$< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}$
$=\frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$
$=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$
$=\frac{1}{4}+\frac{1}{2}-\frac{1}{1000}$
$< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}$
Ta có đpcm.
LT
5 tháng 3 2020
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}\)
=> \(3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{2^{2018}}-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-...-\frac{2019}{4^{2019}}\)
=>3S=\(1+\frac{1}{4}+\frac{1}{4^2}+..+\frac{1}{2^{2018}}-\frac{2019}{4^{2019}}\)
còn lại tự giải nhé
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 \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)
\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)
\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)
\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)
\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)
\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)
\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)
S
0
8 tháng 11 2018
Bài 1 : Ta có : S = 1 + 2 + 22 + 23 + ... + 29
2S = 2(1 + 2 + 22 + 23 + ... + 29)
2S = 2 + 22 + 23 + ... + 210
2S - S = (2 + 22 + 23 + ... + 210) - (1 + 2 + 22 + 23 + ... + 29)
S = 210 - 1 = 28.4 - 1
Vậy S < 5 x 28
PT
1
4 tháng 4 2019
1/2! + 2/3! + 3/4! + ... + 2019/2020!
= (2-1) /2! + (3-1)/3! +(4-1)/4! +....+ (2019-1)/2019! + (2020-1)/2020!
= 2/2! -1/2! + 3/3! -1/3! + 4/4! -1/4! +..........+ 2019/2019! -1/2019! +2020/2020! -1/2020!
= 1/1! -1/2! + 1/2! -1/3! + 1/3! -1/4! +........+ 1/2018! -1/2019! +1/2019! -1/2020!
=1 -1/2020! <1
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2019}{4^{2019}}\)
\(\Rightarrow4S=4(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2019}{4^{2019}})\)
\(\Rightarrow4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}\)
\(\Rightarrow4S-S=(1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}})-(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2019}{4^{2019}})\)
\(\Rightarrow3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}+\frac{2019}{4^{2019}}\)
\(\Rightarrow3S< 1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\left(1\right)\)
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) \(\Rightarrow3S< \frac{4}{3}\)
\(\Rightarrow S< \frac{4}{9}\Rightarrow S< \frac{4}{9}< \frac{4}{8}=\frac{1}{2}\)
\(\Rightarrow S< \frac{1}{2}\left(đpcm\right)\)