B = (1 + 3) + (3²+3³)+.....+(3⁸⁹+3⁹⁰)

  = 4 + 3² .(1 + 3) + .....+3⁹⁰.(1+3)

 = 1 .4 + 3².4 + ..... +3⁹⁰.4

= 4.(1 + 3² + .... +3⁹⁰) chia hết cho 4

\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)

a) Ta có:
\(A=2+2^2+2^3+...+2^{24}\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{22}+2^{23}+2^{24}\right)\)

\(\Rightarrow A=14+...+2^{21}.\left(2+2^2+2^3\right)\)

\(\Rightarrow A=14+...+2^{21}.14\)

\(\Rightarrow A=\left(1+...+2^{21}\right).14⋮14\)( đpcm )

\(A=2+2^2+2^3+...+2^{24}\)

\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{21}+2^{22}+2^{23}+2^{24}\right)\)

\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{21}\left(1+2+2^2+2^3\right)\)

\(\Rightarrow A=2.15+...+2^{21}.15\)

\(\Rightarrow A=15\left(2+...+2^{21}\right)⋮15\left(đpcm\right)\)

b) Mk sửa đề chút là A chia 16 dư 15 nhé

Ta có:

\(A=2+2^2+2^3+...+2^{24}\)

\(\Rightarrow A=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{20}+2^{21}+2^{22}+2^{23}+2^{24}\right)\)

\(\Rightarrow A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{20}\left(1+2+2^2+2^3+2^4\right)\)

\(\Rightarrow A=2.31+...+2^{20}.31\)

\(\Rightarrow A=\left(2+2^{20}\right).31\) 

Vì 31 chia 16 dư 15 nên suy ra đpcm

\(A=2+2^2+2^3+...+2^{2019}\)

   \(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+^9\right)+...+\left(2^{2017}+2^{2018}+2^{2019}\right)\) 

   \(=14+2^4\left(2+2^2+2^3\right)+2^7\left(2+2^2+2^3\right)+...+2^{2017}\left(2+2^2+2^3\right)\) 

   \(=14+2^4.14+2^7.14+...+2^{2017}.14\)

   \(=14\left(1+2^4+2^7+...+2^{2017}\right)⋮14\)

    \(\Rightarrow A⋮14\)

#_ARMY_#

A=2¹+2²+2³+...+2⁶¹+2⁶²+2⁶³

A=(2¹+2²+2³)+...+(2⁶¹+2⁶²+2⁶³⁾

A=14+...+2⁶¹.(2¹+2²+2³)

A=14+...+2⁶¹.14 chia hết cho 14

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