Question

Cho M =1+3+3^2+3^3+...+3^19

Chứng minh M ko chia hết cho 9

M chia hết cho 40

NHANH LÊN NHÉ MN MK SẮP NỘP RỒI Ạ

Answer 1

*\(M=1+3+3^2+3^3+...+\)\(3^{19}=4+3^2+3^3+...+3^{19}\)

Ta có \(3^2⋮3^2=9,3^3⋮3^2=9,...,3^{19}⋮3^2=9\)nhưng \(4⋮̸9\)

=> \(M⋮̸̸9\)

*\(M=1+3+3^2+...+3^{19}\)

        \(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)\)\(+...+\left(3^{16}+3^{17}+3^{18}+3^{19}\right)\)

         \(=40+3^4\left(1+3+3^2+3^3\right)+...+\)\(3^{16}\left(1+3+3^2+3^3\right)\)

         \(=40\left(1+3^4+...+\right)3^{16}⋮40\)

=>\(M⋮40\)

Answer 2

\(a.\) \(M=1+3+3^2+...+3^{19}\)

Ta có: 1+3=4 ko chia hết cho 9, \(3^2⋮9,3^3⋮9,...,3^{19}⋮9\)

\(\Rightarrow\left(1+3\right)+3^2+3^3+...+3^{19}\)ko chia hết cho 9

\(\Rightarrow M\)ko chia hết cho 9. 

Sorry mình ko viết đc dấu ko chia hết vì nó lỗi.

\(b.M=1+3+3^2+3^3+...+3^{19}\)

\(\Rightarrow M=\left(1+3+3^2+3^3\right)+...\)\(+\left(3^{16}+3^{17}+3^{18}+3^{19}\right)\)

\(\Rightarrow M=1\times\left(1+3+3^2+3^3\right)+3^4\)\(\times\left(1+3+3^2+3^3\right)+...+\)\(3^{16}\times\left(1+3+3^2+3^3\right)\)

\(\Rightarrow M=1\times40+3^4\times40+...\)\(3^{16}\times40\)

\(\Rightarrow M=40\times\left(1+3^4+...+3^{16}\right)\)

\(\Rightarrow M⋮40\)

Hok tốt.

Nhớ cho mik đúng nha