\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)

\(2A+A=2^{101}-2\)

\(A=\frac{2^{101}-2}{3}\)

b) tương tự

\(B=\frac{3^{101}+1}{4}\)

A = 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2

   = ( 2¹⁰⁰ + 2⁹⁸ + ... + 2² ) - ( 2⁹⁹ + 2⁹⁷ + ... + 2 )

   = ( 2¹⁰⁰ + 2⁹⁸ + ... + 2² ) - 2( 2⁹⁹ + 2⁹⁷ + ... + 2 ) + ( 2⁹⁹ + 2⁹⁷ + ... + 2 )

   = 2⁹⁹ + 2⁹⁷ + ... + 2 

=> 4A = 2¹⁰³ + 2⁹⁹ + ... + 2³

=> 3A = 2¹⁰³ - 2

=> A = \(\frac{2^{103}-2}{3}\)

a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

\(\Rightarrow A+2A=2^{101}-2\)

  \(A\left(1+2\right)=2^{101}-2\)

  \(A.3=2^{101}-2\)

  \(A=\frac{2^{101}-2}{3}\)

b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3\)

\(3B=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2\)

\(\Rightarrow B+3B=3^{101}-3\)

\(B\left(1+3\right)=3^{101}-3\)

\(4B=3^{101}-3\)

   \(B=\frac{3^{101}-3}{4}\)

Thanks bạn

a, \(A=...\)

=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

=>\(3A=2^{101}-2\)

=>\(A=\frac{2^{101}-2}{3}\)

b, tương tự a \(B=\frac{3^{101}+1}{4}\)

mấy cái này dễ, tự lm ik

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 

a) A =1+3+3²+3³+...+3¹⁰⁰

   3A = 3 + 3²+3³+...+3¹⁰¹

   3A-A=( 3 + 3²+3³+...+3¹⁰¹)-(1+3+3²+3³+...+3¹⁰⁰)

    2A = 3¹⁰¹-1

    A = \(\frac{3^{101}-1}{2}\)

    Thùy An làm sai rùi

a) A=1+3+3^2+...+3^100

3A=3+3^2+....+3^101

3A-A=1+3^101

A=(1+3^101)/2

  a) A=1+3+3²+...+3¹⁰⁰

    3A=    3+3²+...+3¹⁰⁰+3¹⁰¹

3A-A=3¹⁰¹-1

   2A=3¹⁰¹-1

     A=(3¹⁰¹-1):2

4/544

A=(2¹⁰¹-2)/3

B=(3¹⁰¹+1)/4