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

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

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

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

b) tương tự

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

a,tính 2A + A

b,tính 3B+B

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

=> 2A = 2¹⁰¹ - 2¹⁰⁰+2⁹⁹ - 2⁹⁸+...+2³-2²

=> 2A+A= 2¹⁰¹ -2

=> \(A=\frac{2^{101}-2}{3}\)

phần B bn lm tương tự nha!
 

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

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 = \(\frac{2^{103}-2}{3}\)

kho qua de

Ta có:\(F=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

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

\(2F-F=2^{101}-2\)

\(F=2^{101}-2\)

Vậy F=\(2^{101}-2\)

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

Vậy \(A=\dfrac{2^{101}-2}{3}\)

Tính tổng hay rút gọn hay gì bạn?