Trả lời
M=3+3^2+3^3+...+3^100
=(3+3^2)+(3^3+3^4)+...+(3^99+3^100)
=12+3^2.(3^2+3)+...+3^98(3+3^2)
=12+3^2.12+...+3^98.12
=12.(1+3^2+...+3^98) : 12 (: chia hết nha!)
Do 12=3.4:4=>M: 4
a)\(M=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)=4\left(3+3^3+...+3^{99}\right)⋮4\)
\(M=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)=12\left(1+3^2+...+3^{98}\right)⋮12\)
b)\(M=3+3^2+3^3+3^4+...+3^{100}\)
\(=>3M=3^2+3^3+3^4+3^5+...+3^{101}\)
\(=>3M-M=2M=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)
\(=>2M=3^{101}-3\)
Mà \(2M+3=3^n\)nên \(3^{101}-3+3=3^n=>3^{101}=3^n=>n=101\)
Vậy n = 101
a) Ta có M = 3 + 32 + 33 + 34 + ... + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + ... + 398.(3 + 32)
= 12 + 32 . 12 + ... + 398 . 12
= 12.(1 + 32 + ... + 398) (1)
= 4 . 3 . (1 + 32 + ... + 398) \(⋮\)4
Từ (1) ta có : 12.(1 + 32 + ... + 398) \(⋮\)12
b) M = 3 + 32 + 33 + 34 + ... + 3100
3M = 32 + 33 + 34 + 35 ... + 3101
Lấy 3M - M = (32 + 33 + 34 + 35 ... + 3101) - (3 + 32 + 33 + 34 + ... + 3100)
2M = 32 + 33 + 34 + 35 ... + 3101 - 3 - 32 - 33 - 34 - ... - 3100
= 3101 - 3
2M + 3 = 3101 - 3 + 3
2M + 3 = 3101
=> 2M + 3 = 3n = 3101
=> n = 101
