Question

[ホアン イエン ビー]

bài 4:

a) A=2¹⁰⁰-2⁹⁹+2⁹⁸-2⁹⁷+...+2²-2

b) B=3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷+...+3²-3+1

Answer 1

[Xyz OLM]

A = 2¹⁰⁰- 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2

=> 2A =  2¹⁰¹ - 2¹⁰⁰ + 2⁹⁹ - 2⁹⁸ + ... + 2³ - 2² 

Khi đó 2A  + A = (2¹⁰¹ - 2¹⁰⁰ + 2⁹⁹ - 2⁹⁸ + ... + 2³ - 2²) + (2¹⁰⁰- 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2)

=> 3A = 2¹⁰¹ - 2

=> \(A=\frac{2^{201}-2}{3}\)

b) Ta có B = 3¹⁰⁰- 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

=> 3B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸  + ... + 3³ - 3² + 3

Khi đó 3B + B = (3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸  + ... + 3³ - 3² + 3) + (3¹⁰⁰- 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1)

=> 4B = 3¹⁰¹ + 1

=> B = \(\frac{3^{101}+1}{4}\)

Answer 2

[FL.Han_]

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

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

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

<=> \(3A=2^{101}-2\)

=> \(A=\frac{2^{101}-2}{3}\)

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

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

=> \(3A+A=\left(3^{101}-3^{100}+...+3\right)+\left(3^{100}-3^{99}+...+1\right)\)

<=> \(4A=3^{101}+1\)

=> \(A=\frac{3^{101}+1}{4}\)