a.S=3+3²...+3¹⁰⁰

=(3+3²)+...+(3⁹⁹+3¹⁰⁰)

=3(1+3)+...+3⁹⁹(1+3)

=3.4+...+3⁹⁹.4

=4(3+...+3⁹⁹)\(⋮\)4

a: Ta có: \(A=3+3^2+3^3+\cdots+3^{120}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{119}+3^{120}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+\cdots+3^{119}\left(1+3\right)\)

\(=4\left(3+3^3+\cdots+3^{119}\right)\) ⋮4

TA có: \(A=3+3^2+3^3+\cdots+3^{120}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{118}+3^{119}+3^{120}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{118}\left(1+3+3^2\right)\)

\(=13\left(3+3^4+\cdots+3^{118}\right)\) ⋮13

Ta có: \(A=3+3^2+3^3+\cdots+3^{120}\)

\(=\left(3+3^2+\cdots+3^8\right)+\left(3^9+3^{10}+\cdots+3^{16}\right)+\cdots+\left(3^{113}+3^{114}+\cdots+3^{120}\right)\)

\(=3\left(1+3+\cdots+3^7\right)+3^9\left(1+3+\cdots+3^7\right)+\cdots+3^{113}\left(1+3+\cdots+3^7\right)\)

\(=3280\left(3+3^9+\cdots+3^{113}\right)\)

\(=82\cdot40\cdot\left(3+3^9+\cdots+3^{113}\right)\) ⋮82

b: Ta có: \(A=82\cdot40\cdot\left(3+3^9+\cdots+3^{113}\right)\)

\(=10\cdot82\cdot4\cdot\left(3+3^9+\cdots+3^{113}\right)\) ⋮10

=>A có chữ số tận cùng là 0

c:

Sửa đề: 2A+3 là lũy thừa của 3

\(A=3+3^2+3^3+\cdots+3^{120}\)

=>\(3A=3^2+3^3+\cdots+3^{121}\)

=>\(3A-A=3^2+3^3+\cdots+3^{121}-3-3^2-\cdots-3^{120}\)

=>\(2A=3^{121}-3\)

=>\(2A+3=3^{121}\) là lũy thừa của 3

\(S=1-3+3^2-3^3+...+3^{98}-3^{99}=\left(1-3+3^2-3^3\right)+3^4\left(1-3+3^3-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)=\left(-20\right)+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)=\left(-20\right)\left(1+3^4+...+3^{96}\right)⋮20\)

Ta có: \(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)

\(=\left(1-3+3^2-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)\)

\(=-20\cdot\left(1+...+3^{96}\right)⋮20\)

mong các bn giúp mình gấp ạ ^^

2S+1 là lũy thừa của 3

trình bày ra mà kết quả cũng ko đúng

o biết