B = 1/22+1/32+...+1/102
Chứng minh B<1
K
Khách
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H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
TB
2
6 tháng 3 2023
Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
......
\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )
6 tháng 3 2023
Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)
\(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)
.........................
\(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)
=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
=> D < 1 - \(\dfrac{1}{10}\)
=>D < \(\dfrac{9}{10}\)
=> D < \(\dfrac{10}{10}\)
Vậy D < 1
NT
1
9 tháng 7 2023
A=1/2^2+1/3^2+...+1/10^2
=>A<1-1/2+1/2-1/3+...+1/9-1/10=1-1/10<1
PT
1
NN
14 tháng 8 2023
1.
a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)
\(2A=2+2^2+2^3+....+2^{2008}\)
b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)
\(=2^{2008}-1\) (bạn xem lại đề)
2.
\(A=1+3+3^1+3^2+...+3^7\)
a. \(2A=2+2.3+2.3^2+...+2.3^7\)
b.\(3A=3+3^2+3^3+...+3^8\)
\(2A=3^8-1\)
\(=>A=\dfrac{2^8-1}{2}\)
3
.\(B=1+3+3^2+..+3^{2006}\)
a. \(3B=3+3^2+3^3+...+3^{2007}\)
b. \(3B-B=2^{2007}-1\)
\(B=\dfrac{2^{2007}-1}{2}\)
4.
Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)
a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)
b.\(4C-C=4^7-1\)
\(C=\dfrac{4^7-1}{3}\)
5.
\(S=1+2+2^2+2^3+...+2^{2017}\)
\(2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(S=2^{2018}-1\)
14 tháng 8 2023
4:
a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6
=>4*C=4+4^2+...+4^7
b: 4*C=4+4^2+...+4^7
C=1+4+...+4^6
=>3C=4^7-1
=>\(C=\dfrac{4^7-1}{3}\)
5:
2S=2+2^2+2^3+...+2^2018
=>2S-S=2^2018-1
=>S=2^2018-1
DT
0
HD
1
29 tháng 6 2015
a) gọi dãy trên là A
A<1/100+1/100+...+1/100(100 lần 1/100 vì từ 101đến 200 có 100 số)
A<1*100/100=1
A<1
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(.....\)
\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)
\(\Rightarrow B< 1\left(dpcm\right)\)
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(B< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)
\(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(B< 1-\dfrac{1}{10}\)
\(B< \dfrac{9}{10}< 1\)
Vậy \(B< 1\)