chứng tỏ: 1/3+2/3^2+......+2019/362019<0,75
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OT
1
15 tháng 2 2020
Đặt A=\(\frac{1}{3}+\frac{2}{3^2}+.....+\frac{2019}{3^{2019}}\)
3A=\(1+\frac{2}{3}+.....+\frac{2019}{3^{2018}}\)
3A - A = \(\left(1+\frac{2}{3}+...+\frac{2018}{3^{2017}}+\frac{2019}{3^{2018}}\right)\) -\(\left(\frac{1}{3}+....+\frac{2017}{3^{2017}}+\frac{2018}{3^{2018}}+\frac{2019}{3^{2019}}\right)\)
2A = \(1+\frac{1}{3}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt B=\(1+\frac{1}{3}+....+\frac{1}{3^{2018}}\)
3B =\(3+1+....+\frac{1}{3^{2017}}\)
3B - B=\(\left(3+1+....+\frac{1}{3^{2017}}\right)\)-\(\left(1+\frac{1}{3}+...+\frac{1}{3^{2018}}\right)\)
2B =\(3-\frac{1}{3^{2018}}\)
Ta có:2A= B - \(\frac{2019}{3^{2019}}\)
4A = 2B -\(\frac{2.2019}{3^{2019}}\)
4A=\(\left(3-\frac{1}{3^{2018}}\right)\)-\(\frac{2.2019}{3^{2019}}\)
A=\(\frac{3}{4}-\frac{1}{3^{2018}.4}-\frac{2019}{3^{2019}.2}\)<\(\frac{3}{4}\)=0,75
Suy ra :\(\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)< 0,75 (đpcm)
LG
1
DH
18 tháng 2 2020
Đặt: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2019}{3^{2018}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2018}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2018}}}{2}\)
Thay vào \(2A\Rightarrow2A=1+\frac{\left(1-\frac{1}{3^{2018}}\right)}{2}-\frac{2019}{3^{2019}}\)
\(=1+\frac{1}{2}-\frac{1}{2.3^{2018}}-\frac{2019}{3^{2019}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< 0,75\left(đpcm\right)\)
T
0
15 tháng 4 2019
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)
=>\(3A=1+\frac{2}{3}+...+\frac{2019}{3^{2018}}\)
=>\(2A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{2018}}\)
=>\(2B=3-\frac{1}{3^{2018}}\)=>\(B=\frac{3-\frac{1}{3^{2018}}}{2}\)
=>\(2A=\frac{3-\frac{1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}=\frac{\frac{3^{2019}-1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}\)
\(=\frac{3^{2019}-1}{3^{2018}.2}-\frac{2019}{3^{2019}}=\frac{3\left(3^{2019}-1\right)-2019.2}{3^{2019}.2}\)
15 tháng 4 2019
Nhầm tí
dòng thứ 2 từ dưới lên cm bé hơn 0,75 luôn nhá
25 tháng 3 2017
0,75 = \(\dfrac{3}{4}\)
Ta có: \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... +\(\dfrac{1}{2000.2001}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2000}\) - \(\dfrac{1}{2001}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2}\) - \(\dfrac{1}{2001}\).
Vì \(\dfrac{1}{2}\) < \(\dfrac{3}{4}\) nên \(\dfrac{1}{2}\) - \(\dfrac{1}{2001}\) < \(\dfrac{3}{4}\).
Vậy \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{3}{4}\).
NN
0
T
23 tháng 5 2018
Làm theo cách của Trắng nha ,
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\left(Đpcm\right)\)
23 tháng 5 2018
Ta có: \(\frac{1}{2^2}=\frac{1}{2^2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...................
\(\frac{1}{2019^2}< \frac{1}{2018.2019}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(=\frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2019}\)
\(=\frac{1}{4}+\frac{2}{4}-\frac{1}{2019}\)
\(=\frac{3}{4}-\frac{1}{2019}\)\(< \frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2019^2}< \frac{3}{4}\)
Điều phải chứng minh
22 tháng 5 2018
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2019^2}\)
\(\Rightarrow A=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2019}\right)\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\)
9 tháng 4 2019
Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)
\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)
\(\Rightarrow S< 2\) ( ĐPCM )