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 \(C=\dfrac{4^{101}-1}{3}\)

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

2D = 2¹⁰¹ - 2¹⁰⁰ + 2⁹⁹ -...+2

2D+D= 2¹⁰¹+1

D=...

Bạn tự tính nhé nhớ k cho mình đấy

\(\frac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)

\(=\frac{\left(101+1\right).100:2}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)

\(=\frac{5050}{1+1+...+1+1}\)(51 chữ số 1)

= \(\frac{5050}{51}\)

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.

Lời giải:

a) \(A=1+3+3^2+3^3+...+3^{100}\)

\(\Rightarrow 3A=3+3^2+3^3+...+3^{101}\)

Trừ theo vế:
\(\Rightarrow 3A-A=(3+3^2+3^3+..+3^{101})-(1+3+3^2+...+3^{100})\)

\(2A=3^{101}-1\Rightarrow A=\frac{3^{101}-1}{2}\)

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

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

Cộng theo vế:

\(\Rightarrow B+2B=2^{201}-2\)

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

c) Ta có:

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

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

Cộng theo vế:

\(C+3C=(3^{100}-3^{99}+3^{98}-....+3^2-3+1)+(3^{101}-3^{100}+3^{99}-....+3^3-3^2+3)\)

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

a: \(3A=3+3^2+...+3^{101}\)

\(\Leftrightarrow2A=3^{101}-1\)

hay \(A=\dfrac{3^{101}-1}{2}\)

b: \(2B=2^{101}-2^{100}+...+2^3-2^2\)

\(\Leftrightarrow3B=2^{101}-2\)

hay \(B=\dfrac{2^{101}-2}{3}\)

c: \(3C=3^{101}-3^{100}+....+3^3-3^2+3\)

=>\(4C=3^{101}+1\)

hay \(C=\dfrac{3^{101}+1}{4}\)

Đặt A = 2 ^ 100 + 2 ^ 99 + 2 ^ 98 + ... + 2 ^ 2 + 2 ^ 1

 2A    = 2 ^ 101 + 2 ^ 100 + 2 ^ 99 + ... + 2 ^ 3 + 2 ^ 2

2A - A = ( 2 ^ 101 + 2 ^ 100 + 2 ^ 99 + ... + 2 ^ 3 + 2 ^ 2 )

           -  ( 2 ^ 100 + 2 ^ 99 + 2 ^ 98 + ... + 2 ^ 2 + 2 ^ 1 )

 A        = 2 ^ 101 - 2

\(A=2^{100}+2^{99}+2^{98^{ }}+...+2^2+2^1\)

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

\(2A=2^{101}+2^{100}+...+2^2+2^1\)

\(A=2A-A\)

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

Viết biểu thức A thành:
\(A=\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}-2^{97}-2\right)=M-N\)
Ta có \(M=2^{100}+2^{98}+...+2^2\)
\(\Rightarrow2^2M=2^{102}+2^{100}+...+2^4\)
\(\Rightarrow4M-M=2^{102}-2^2\Rightarrow M=\frac{3^{102}-2}{3}\)
Tương tự với \(N=2^{99}-2^{97}-...-2=\frac{3^{101}-3^{100}+2}{3}\)
NHư vậy \(M-N=\frac{2^{102}-2-2^{101}+2^{100}-2}{3}=\frac{3^{101}.34}{3}\)