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) 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 = 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!
 

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

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

\(2A-A=\left(2^2+2^3+....+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

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

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

=>2A=2^2+2^3+2^4+...+2^100+2^101

=> 2A - A =(2^2+2^3+2^4+...+2^100+2^101)-(2+2^2+2^3+...+2^99+2^100)

=>A = 2^101-2

A = 2 + 2^2 + 2^3 + 2^4 +...+ 2^99 + 2^100

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

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

A = 2^101 - 2

\(B=1^2+2^2+\cdot\cdot\cdot+100^2\)

\(\Rightarrow B=1\cdot\left(2-1\right)+2\cdot\left(3-1\right)+\cdot\cdot\cdot+100\cdot\left(101-1\right)\)

\(\Rightarrow B=\left(1\cdot2+2\cdot3+\cdot\cdot\cdot+100\cdot101\right)-\left(1+2+\cdot\cdot\cdot+100\right)\)

Đặt A = 1.2 + 2.3 + ... + 100.101

\(\Rightarrow3A=1\cdot2\cdot3+2\cdot3\cdot3+\cdot\cdot\cdot+100\cdot101\cdot3\)

\(\Rightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+\cdot\cdot\cdot+100\cdot101\cdot\left(102-99\right)\)

\(\Rightarrow3A=\left(1\cdot2\cdot3+\cdot\cdot\cdot+100\cdot101\cdot102\right)-\left(1\cdot2\cdot3+\cdot\cdot\cdot+99\cdot100\cdot101\right)\)

\(\Rightarrow3A=100\cdot101\cdot102\)

\(\Rightarrow A=100\cdot101\cdot34\)

\(\Rightarrow A=343400\)

\(\Rightarrow B=A-\left(1+2+\cdot\cdot\cdot+100\right)\)

\(\Rightarrow B=343400-\frac{101\cdot100}{2}\)

\(\Rightarrow B=343400-101\cdot50\)

\(\Rightarrow B=343400-5050\)

\(\Rightarrow B=338350\)

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

\(3A=3^2+3^3+3^4+...+3^{101}+3^{102}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{101}+3^{102}\right)-\left(3+3^2+3^3+...+3^{100}+3^{101}\right)\)

\(2A=3^{102}-3\)

\(A=\frac{3^{102}-3}{2}\)

Tớ chỉ làm được câu A thôi, bạn thông cảm. Với lại tớ không chắc đúng đâu.

=))

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 = 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰ . (1)

\(\Rightarrow2A=2+2^2+2^3+...+2^{101}\) (2)

Trừ 2 vế của (1) và (2) cho nhau được \(A=2^{101}-1\)

Ta có: A = 1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰.

        2A = 2 (1 + 2 + 2² + 2³ + ... + 2⁹⁹ + 2¹⁰⁰)

             = \(2\cdot1+2\cdot2+2\cdot2^2+2\cdot2^3+...+2\cdot2^{99}+2\cdot2^{100}.\)

       2A  = \(2+2^2+2^3+2^4+...+2^{100}+2^{101}.\)

  2A - A  = \(\left(2+2^2+2^3+2^4+...+2^{100}+2^{101}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

         A  = \(2^{101}-1\).

          Vậy A = 2¹⁰¹ - 1.