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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)
Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)
\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)
\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)
Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)
Vậy A < 2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1+2+3+...+99 = 99.50
A= 13 + 23 + ... + 993 = (13 + 993 ) + (23 + 982) + ... + 503 chia hết cho 50
A = 13 + 23 + ... + 993 = (13 + 983 ) + (23 + 972) + ... + 993 chia hết cho 99
=> A chia hết cho 99.50 => A chia hết cho B => số dư = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(3^{32}-1=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C1: 2A = (3-1 )(3+1)....(3^16)
Theo hằng đẳng thức thì a^2-b^2 thì 2A = 3^32 - 1 => A = (3^32-1) /2<B
C2 biến đổi B
B = 3^32 - 1 = (3^16- 1)(3^16 + 1)=(3^8 - 1)(3^8 + 1)= ... = 2(3+1)(3^2+1)..(3^16+1)>A
![](https://rs.olm.vn/images/avt/0.png?1311)
B=(3+1).(3^2+1).(3^4+1).(3^8+1).(3^16 +1).2
=>B=2.(3+1)(32+1)(34+1)(38+1).(316+1)
=(3-1)(3+1)(32+1)(34+1)(38+1).(316+1)
=(32-1)(32+1)(34+1)(38+1).(316+1)
=(34-1)(34+1)(38+1).(316+1)
=(38-1)(38+1).(316+1)
=(316-1)(316+1)
=332-1=A
Vậy A=B
\(B=2.\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^4-1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^8-1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right).\left(3^{16}+1\right)\)
\(=3^{32}-1\)
Vậy A = B = 332 - 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\)
\(=\frac{1}{2}\left(3^8-1\right)\)
Vậy A < B
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(2A=\left(3^8-1\right)\)
\(A=\frac{3^8-1}{2}< B\)
A= 1/3+1/3^2+1/3^3+...+1/3^99
=> 3A=1+1/3+1/3^2+...+1/3^98
Vậy 2A= 3A-A= 1-1/3^99
=> A= 1/2 -1/ 2.3^99
=> A < 1/2
Mik giải ngắn gọn thôi nha!