E= 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ + 3⁹⁷ + .... + 3² - 3 + 1

3A = 3¹⁰¹- 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + .... - 3² + 3 

3A + A = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ...... - 3² + 3 + 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - .......- 3² - 3 + 1 

 4A      = 3¹⁰¹ + 1

=> A = \(\frac{3^{101}+1}{4}\) 

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 \(C=\dfrac{3^{101}+1}{4}\)

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=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^ 201 -2

=>A=\(\frac{2^{201}-2}{3}\)

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=3^ 101 -3^100+3^ 99 -3 ^98+...+3 ^3 -3 ^2+3+3 ^100 -3^ 99+3^ 98 -3^ 97+....+3 ^2 -3+1

=>4B=3 ^101+1

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

\(C=1\cdot99+2\cdot98+3\cdot97+...+98\cdot2+99\cdot1\)

\(C=\left(1+2+3+...+98+99\right)\left(99+98+...+3+2+1\right)\)

Mà    \(\left(1+2+3+...+98+99\right)=\left(99+98+...+3+2+1\right)\)

\(\Rightarrow C=\left(1+2+3+...+98+99\right)^2\)

Tính \(1+2+3+...+98+99\)

\(=\left(99+1\right)+\left(98+2\right)+\left(97+3\right)+.....\)

\(=100\cdot\frac{99}{2}=4950\)

Có \(C=\left(1+2+3+...+98+99\right)^2\)

\(\Rightarrow C=4950^2\)

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

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

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

  \(A\left(1+2\right)=2^{101}-2\)

  \(A.3=2^{101}-2\)

  \(A=\frac{2^{101}-2}{3}\)

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

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

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

\(B\left(1+3\right)=3^{101}-3\)

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

   \(B=\frac{3^{101}-3}{4}\)

Thanks bạn

a, \(A=...\)

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

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

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

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

b, tương tự a \(B=\frac{3^{101}+1}{4}\)