= ? bao nhiêu vậy các bạn giúp mình gấp nheqq 

50 

tick minh nha

các bạn trình bày cách làm mình với ????

\(\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}+\frac{99}{1}}\)

Xét M - 99 + 98 = \(\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}\)

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

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

\(\Rightarrow\frac{T}{M}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)

Vế A

Ta có : A =

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

=> 2A + A = 3A = \(2^{100}-2\Rightarrow A=\dfrac{2^{100}-2}{3}\)

=================

B làm tương tự , nhân 3 lên rồi cộng lại là ra

b) B = 2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ...+ 2² - 2

=> B x 2 = 2^{101 -} 2¹⁰⁰ + 2⁹⁹ -  2⁹⁸  + ...2³ - 2²

=> B x 2 + B = (2^{101 -} 2¹⁰⁰ + 2⁹⁹ -  2⁹⁸  + ...2³ - 2² ) + (2¹⁰⁰ - 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ...+ 2² - 2)

  <=>  B x 3 = 2¹⁰¹ - 2 = 2. ( 2⁹⁹ - 1)

=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)

Phần c) Làm tương tự Lấy C x 3 rồi + với C.

a, \(A=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)

\(\Rightarrow3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)

\(\Rightarrow4A=3^{101}+1\)

\(\Rightarrow A=\dfrac{3^{101}+1}{4}\)

Vậy...

b, tương tự

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

\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+..+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)\(\Rightarrow3A=2^{101}-2\)

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

Có: \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)

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

=>\(3B+B=\left(3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\right)+\left(3^{100}-3^{99}+3^{98}-3^{97}+...-3+1\right)\)

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

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

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)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

b)

B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

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!