chung minh rang 1/22+1/32+1/42+.......+1/20122<1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NB
0
PM
0
DT
1
5 tháng 3 2017
ta có :1+(1/2+1/2+........+1/32)
1+(1/2*3*4*5......*32)
=>1/2*3*4*....*32<1
vậy 1+1/2+1/3+1/4+........+1/32<3
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
NH
0
NI
0
\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3};...;\frac{1}{2012^2}=\frac{1}{2012.2012}<\frac{1}{2011.2012}\)
\(=>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{2011.2012}\)
\(=>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{2011}-\frac{1}{2012}=\frac{1}{1}-\frac{1}{2012}=\frac{2011}{2012}<1\)
=>đpcm