giúp minh voi

\(A=1+2^1+2^2+...+2^{2017}\)

\(2A=2+2^2+2^3+...+2^{2018}\)

\(2A-A=2^{2018}-1hayA=2^{2018}-1\)

2; 3 tuong tu

1) A = 1 + 2 + 2² + 2³ + .... + 2²⁰¹⁸

2A = 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰¹⁹

2A - A = ( 2 + 2² + 2³ + 2⁴ + ..... + 2²⁰¹⁹ ) - ( 1 + 2 + 2² + 2³ + .... + 2²⁰¹⁸ )

Vậy A = 2²⁰¹⁹ - 1

2) B = 1 + 3 + 3² + 3³ + ..... + 3²⁰¹⁸

3A = 3 + 3² + 3³ + ...... + 3²⁰¹⁹

3A - A = ( 3 + 3² + 3³ + ...... + 3²⁰¹⁹ ) - ( 1 + 3 + 3² + 3³ + ..... + 3²⁰¹⁸ )

2A = 3²⁰¹⁹ - 1

Vậy A = ( 3²⁰¹⁹ - 1 ) : 2

3) C = 1 + 4 + 4² + 4³ + ...... + 4²⁰¹⁸

4A = 4 + 4² + 4³ + ...... + 42019

4A - A = ( 4 + 4² + 4³ + ...... + 4²⁰¹⁹ ) - ( 1 + 4 + 4² + 4³ + ...... + 4²⁰¹⁸ )

3A = 4²⁰¹⁹ - 1

Vậy A = ( 4²⁰¹⁹ - 1 ) : 3

1x2x3x...2018x2019 - 1x2x3x..2018 - 1x2x3x4x...x2017x2018² 

= 1x2x3x...x2018x(2019 - 1 - 2018)

= 1x2x3x...x2018x0

= 0

đặt A=1+4+4^2+4^3+...+4^2018

    B=1+2+2^2+2^3+...+2^2018

A=1+4+4^2+4^3+...+4^2018

4A=4+4^2+4^3+...+4^2019

4A-A=(4+4^2+4^3+...+4^2019)-(1+4+4^2+4^3+...+4^2018)

3A=4^2019-1

A=(4^2019)/3

B=1+2+2^2+2^3+...+2^2018

2B=2+2^2+2^3+...+2^2019

2B-B=(2+2^2+2^3+...+2^2019)-(1+2+2^2+2^3+...+2^2018)

B=2^2019-1

=>(1+4+4^2+4^3+...+4^2018)/(1+2+2^2+2^3+...+2^2018) =A/B=(4^2019-1)/3/(2^2019-1)

=(4^2019-1)/(3.2^2019-3)

Vậy ...............................

+) Gọi A là tổng của dãy số: 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018.
+) Số số hạng của A là:
A = (2018 - 1) : 1 + 1 = 2018.
+) Tổng A là: (2018 + 1). 2018 : 1 = 4074342.
Vậy, A = 4074342 (hay 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018 = 4074342). 

Ah bạn à chia 2 mà ._. Nhưng mà cảm ơn

a>b

K CHO MÌNH NHÉ

cái này là khi chiều mới thi nầy

Giải:

Ta có:A=1.2+2.3+3.4+...+2017.2018

         3A=1.2.3 2.3.3+...+2017.2018.3

             =1.2.(3-0)+2.3.(4-1)+...+2017.2018.(2019-2016)

             =1.2.3+2.3.4+...+2017.2018.2019-1.2.0-2.3.1-...-2017.2018.1016

             =2017.2018.2019-1.2.0

             =2017.2018.2019

           =>A=2017.2018.2019/3=2018.(2017.2019)/3

            Và B=2018³/3=2018.2018.2018/3=2018.(2018.2018)/3

 Lại có: 2017.2019=2017.(2018+1)=2017.2018+2017

            2018.2018=(2017+1).2018=2017.2018+2018

Mà 2017.2018+2017<2017.2018+2018 =>2017.2019<2018.2018

    =>2018.(2017.2019)<2018.(2018.2018)

   =>A=2018.(2017.2019)/3<2018.(2018.2018)/3=B

   =>A<B

  Vậy A<B

Chúc Công Chúa Bloom học giỏi!!!

\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)

\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)

\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)

Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)

=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)

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