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

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

Ta có:

\(3^{32}-1=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow A< B\)

Vậy A = B

tích mik nha mik se cho lời giải

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 

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

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)

\(=\frac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\)

\(=\frac{1}{2}\left(3^8-1\right)\)

Vậy A < B

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\)

\(2A=\left(3^8-1\right)\)

\(A=\frac{3^8-1}{2}< B\)