Ta có: 3¹+3²+3³+…+3⁹⁹+3¹⁰⁰

=(3¹+3²)+(3³+3⁴)+…+(3⁹⁹+3¹⁰⁰)

=3.(1+3)+3³.(1+3)+…+3⁹⁹.(1+3)

=3.4+3³.4+…+3⁹⁹.4

=(3+3³+…+3⁹⁹).4 chia hết cho 4

=>3¹+3²+3³+…+3⁹⁹+3¹⁰⁰ chia hết cho 4

Đặt   \(A=3+3^2+3^3+...+3^{99}+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^{ 3}\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+...+3^{99}\right)\)

\(=4\left(3+3^3+...+3^{99}\right)\)

Vì 4 chia hết cho 4 nên \(4\left(3+3^3+...+3^{99}\right)\)

Vậy A chia hết cho 4

a) Ta có : M = 3 + 3²  + 3³ + ... + 3¹⁰⁰

=> M = (3 + 3²) + (3³ + 3⁴) + ... + (3⁹⁹ + 3¹⁰⁰)

=> M = 12 + 3²(3 + 3²) + ... + 3⁹⁸(3 + 3²)

=> M = 12 + 3².12 + ... + 3⁹⁸.12

=> M = 12(1 + 3² + ... + 3⁹⁸) \(⋮\)12

Do 12 = 3 . 4 \(⋮\)4 => M \(⋮\)4

b) Ta có: 2m + 3 = 3

=> 2m = 3 - 3

=> 2m = 0

=> m = 0 : 2

=> m = 0

Đặt A = 1/3 + 2/3² + 3/3³ + 4/3^4 + ... + 100/3^100 

=> 3A= 1 + 2/3 + 3/3² + 4/3³ + .... + 100/3^99 

=> 3A-A = 1 + (2/3 - 1/3) + (3/3² - 2/3²) +...+ (100/3^99 - 99/3^99) - 100/3^100

=> 2A= 1+ 1/3 + 1/3² + 1/3³ +...+ 1/3^99 - 100/3^100

Đặt B = 1/3 + 1/3² + 1/3³ +...+ 1/3^99 

=> 3B = 1 + 1/3 + 1/3² + 1/3³ +...+ 1/3^98

=> 2B = 1 - 1/3^99 => B = (1 - 1/3^99)/2

Thay vào 2A => 2A= 1+ 1/2 - 1/(2x3^99) - 100/3^100 < 1+ 1/2 = 3/2 

=> A < 3/4

....

Ý Trước

\(A=2^{100}+2^{101}+2^{102}+...+2^{107}\)

\(A=2^{100}\left(1+2\right)+2^{102}\left(1+2\right)+...+2^{106}\left(1+2\right)\)

\(A=2^{100}.3+2^{102}.3+...+2^{106}.3\)

\(A=3\left(2^{100}+2^{102}+...+2^{106}\right)⋮3\)