A= 1/32 + 3/34+5/36+…..+99/3100 CMR A < 5/32
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
S
1
S
0
TD
0
AD
0
VG
1
MV
0
P
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
5 tháng 3 2021
\(A=1+3^2+3^4+...+3^{102}\)
\(9A=3^2+3^4+...+3^{102}+3^{104}\)
\(\Rightarrow9A-A=3^{104}-1\)
\(\Rightarrow8A=3^{104}-1\)
\(\Rightarrow A=\dfrac{3^{104}-1}{8}\)
KJ
1
Q
13 tháng 12 2021
Tham khảo
Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)
3A = 3+32+33+...+3100+31013+32+33+...+3100+3101
Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)
2A = 3101−13101−1
⇒⇒ A = 3101−123101−12
Vậy A = 3101−12
A=\(\frac{1}{3^2}+\frac{3}{3^4}+\frac{5}{3^6}+\cdots+\frac{99}{3^{100}}\)
nhân biểu thức A với 9
\(9A=9\left(\frac{1}{3^2}+\frac{3}{3^4}+\frac{5}{3^6}+\cdots+\frac{99}{3^{100}}\right)\)
\(9A=1+\frac{3}{3^2}+\frac{5}{3^4}+\cdots+\frac{99}{3^{98}}\)
\(9A-A=\left(1+\frac{3}{3^2}+\frac{5}{3^4}+\cdots+\frac{99}{3^{98}}\right)-\left(\frac{1}{3^2}+\frac{3}{3^4}+\frac{5}{3^6}+\cdots+\frac{99}{3^{100}}\right)\) \(8A=1+\left(\frac{3}{3^2}-\frac{1}{3^2}\right)+\left(\frac{5}{3^4}-\frac{3}{3^4}\right)+\cdots+\left(\frac{99}{3^{98}}-\frac{97}{3^{98}}\right)-\frac{99}{3^{100}}\)
8A=\(1+\frac{2}{3^2}+\frac{2}{3^4}+\cdots+\frac{2}{3^{98}}-\frac{99}{3^{100}}\)
8A=\(1+2\left(\frac{1}{3^2}+\frac{1}{3^4}+\cdots+\frac{1}{3^{96}}\right)-\frac{99}{3^{100}}\)
đặt B=\(\frac{1}{3^2}+\frac{1}{3^4}+\cdots+\frac{1}{3^{98}}\)
=> 9B=\(1+\frac{1}{3^2}+\frac{1}{3^4}+\cdots+\frac{1}{3^{96}}\)
\(9B-B=\left(1+\frac{1}{3^2}+\frac{1}{3^4}+\cdots+\frac{1}{3^{96}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^4}+..+\frac{1}{3^{98}}\right)=1-\frac{1}{3^{98}}\)
=> \(B=\frac{\left(1-\frac{1}{3^{98}}\right)}{8}\)
thay ngược lại vào biểu thức 8A ta có:
\(8A=1+2\left(\frac{\left(1-\frac{1}{3^{98}}\right)}{8}\right)-\frac{99}{3^{100}}\)
8A=\(1+\left(\frac{\left(1-\frac{1}{3^{98}}\right)}{4}\right)-\frac{99}{3^{100}}\)
8A=\(1+\frac14-\frac{1}{4\cdot3^{98}}-\frac{99}{3^{100}}\)
8A=\(\frac54-\left(\frac{1}{4\cdot3^{98}}+\frac{99}{3^{100}}\right)\)
=> \(8A<\frac54\)
=> \(A<\frac{5}{4\cdot8}=\frac{5}{32}\left(đpcm\right)\)