a: \(2A=2-2^2+2^3-...-2^{2002}+2^{2003}\)

\(\Leftrightarrow3A=2^{2003}+1\)

hay \(A=\dfrac{2^{2003}+1}{3}\)

b: \(3B=3-3^2+3^4-3^4+...+3^{99}-3^{100}\)

\(\Leftrightarrow4B=-3^{100}+1\)

hay \(B=\dfrac{-3^{100}+1}{4}\)

AI MUỐN KẾT BẠN VỚI MÌNH KHÔNG VẬY ?

ố 29 phút trước tui làm gì lên

Hỏi từng chút có được ko dzậy

a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)

\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B

=>B/A=1/100

b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)

\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)

\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)

\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)

=>A/B=25

a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)

\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)

\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)

Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)

b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)

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

\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)

Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)

c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)

\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)

\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)

\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)

\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)​

\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)

a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)

\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)

\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)

\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)

\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)

Vậy \(A:B=1.\)

b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)

\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)

\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)

\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)

Vậy \(A:B=1.\)

c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)

\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)

\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)

\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)

\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)

Vậy \(A:B=1.\)

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

Bài 1: 

a: \(2A=2^{101}+2^{100}+...+2^2+2\)

\(\Leftrightarrow A=2^{100}-1\)

b: \(3B=3^{101}+3^{100}+...+3^2+3\)

\(\Leftrightarrow2B=3^{100}-1\)

hay \(B=\dfrac{3^{100}-1}{2}\)

c: \(4C=4^{101}+4^{100}+...+4^2+4\)

\(\Leftrightarrow3C=4^{101}-1\)

hay \(C=\dfrac{4^{101}-1}{3}\)

đề câu b sai rồi ,như thế này mới đúng :

b .1²+2²+3²+...+98²

                                                          giải:

        ta có: a=1.2+2.3+3.4+...+98.99

                  b=1²+2²+3²+...+98⁸

                 .=1.1+2.2+3.3+...+98.98

                  =(1.2-1.1)+(2.3-2.2)+(3.4-3.3)+...+(98.99-98.98)

                  =1(2-1)+2(3-2)+3(4-3)+...+98(99-98)

                  =1.1+2.1+3.1+...+98.1

                  =1+2+3+...+98

                  =[98.(98+1)] / 2 

                  = 98 .99 / 2

                  =    4851

                  vậy :a-b = 4851

a = 1 . 2 + 2 . 3 + 3. 4 + .... + 98 . 99 

3a = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + .... + 98 . 99 . ( 100 - 97 )

3a = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + ... 98 . 99 . 100 - 97 . 98 . 99

3a = 98 . 99 . 100 = 970200

a = 323400

a = 1 . 2 + 2 . 3 + 3. 4 + .... + 98 . 99 

3a = 1 . 2 . 3 + 2 . 3 . ( 4 - 1 ) + 3 . 4 . ( 5 - 2 ) + .... + 98 . 99 . ( 100 - 97 )

3a = 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + ... 98 . 99 . 100 - 97 . 98 . 99

3a = 98 . 99 . 100 = 970200

a = 323400

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