Question

Bài 2 

A=3+3²+...+³¹⁰⁰

a,tìm số tự nhiên n để 2.A+3=3ⁿ

chứng tỏ A⋮12

Answer 1

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

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

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

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

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

mà \(2A+3=3^n\)

nên n=101

b: \(A=3+3^2+...+3^{100}\)

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

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{98}\right)⋮12\)