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![](https://rs.olm.vn/images/avt/0.png?1311)
ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2018}{3^{2018}}\)(1)
\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{2018}{3^{2019}}\)(2)
Lấy (1) trừ (2) theo vế ta có :
\(A-\frac{1}{3}A=\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2018}{3^{2018}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{2018}{3^{2019}}\right)\)
\(\Rightarrow\frac{2}{3}A=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\right)-\frac{2018}{3^{2019}}\)
Đặt B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
=> 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
Lấy 3B trừ B theo vế ta có :
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\right)\)
=> 2B = \(1-\frac{1}{3^{2018}}\)
=> \(B=\frac{1}{2}-\frac{1}{3^{2018}.2}\)
Khi đó : \(\frac{2}{3}A=\frac{1}{2}-\frac{1}{3^{2018}.2}-\frac{2018}{3^{2019}}\)
\(A=\left(\frac{1}{2}-\frac{1}{3^{2018}.2}-\frac{2018}{3^{2019}}\right):\frac{2}{3}=\frac{3}{4}-\frac{1}{3^{2017}.4}-\frac{1009}{3^{2018}}=\frac{3}{4}-\left(\frac{1}{3^{2017}.\left(3+1\right)}+\frac{1009}{3^{2018}}\right)\)
\(=\frac{3}{4}-\left(\frac{1}{3^{2018}}+\frac{1}{3^{2017}}-\frac{1009}{3^{2018}}\right)=\frac{3}{4}-\left(\frac{1}{3^{2017}}-\frac{336}{3^{2017}}\right)=\frac{3}{4}+\frac{335}{3^{2017}}\)
Vì A > 0 (1)
Mặt khác\(\frac{335}{3^{2017}}< \frac{335}{1340}< \frac{1}{4}\)
=> \(\frac{335}{3^{2017}}< \frac{1}{4}\Rightarrow\frac{3}{4}+\frac{335}{3^{2017}}< \frac{1}{4}+\frac{3}{4}\Rightarrow A< 1\)(2)
Từ (1) và (2) => 0 < A < 1
=> A không phải là số nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
=>2B-B=\(1-\frac{1}{2^{99}}\)
mà 1/2^99>0 nên B<1 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
C = 1/3 + 1/3^2 + 1/3^3 + ... =1/3^99
=> C = 1/3^99 = 1/(3^99)
=> C < 1/2 (đpcm)
2A=2^101-2^100+2^98+...+2^3-2^2
3A = 2A + A
3A = 2^101 - 2 ( Cứ tính là ra , âm vs dương triệt tiêu )
A = (2^101-2) :3
B tăng tự
![](https://rs.olm.vn/images/avt/0.png?1311)
A= 20-21+22-23+24-25...+22018
A=20-21+22-23(20-21+22)+26(...)+...22016(20+21+22)
xet thay 20-21+22 chia het cho 3
thi toan bo day co 673 cap chia het cho 3
k please