bài 1

chứng minh chia hết cho 3 nè

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

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

s=\(2.\left(1+2\right)+2^2.\left(1+2\right)+...+2^{99}.\left(1+2\right)\)

s=\(2.3+2^2.3+...+2^{99}.3\)

s=\(3.\left(2+2^2+...+2^{99}\right)\)chia hết cho 3 => s chia hết cho 3(đpcm)

chứng minh chia hết cho 5

s=\(\left(2+2^2+2^3+2^4\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

s=\(2.\left(1+2+4+8\right)+...+2^{97}.\left(1+2+4+8\right)\)

s=\(2.15+...+2^{97}.15\)

s=\(15.\left(2+...+2^{97}\right)\)chia hết cho 5=> s chia hết cho 5

mong là có thể giúp được bạn

tui ko  bit

S=5+5^2+5^3+....+5^96= 
= 5+5^2+5^3+ 5^4+5^5+5^6....+ +5^91 + 5^92+5^93 +5^94 +5^95 +5^96 
=(5+5^2+5^3+ 5^4+5^5+5^6)(1+5^6 + ... +5^90)= 
=5* 126*31*(1+5^6 + ... +5^90)= 5* 126*31*(1+5^4 + ... +5^90) chia hết cho 126 

Bạn gộp 6 số lại là được 

= 5^2017( 1+5-5^2)

=5^2017. (-19) chia hết cho 19

\(5^{2017}+5^{2018}-5^{2019}=5^{2017}\left(1+5-5^2\right)=5^{2017}\left(-19\right)⋮19\)

5²⁰¹⁷ + 5²⁰¹⁸ + 5²⁰¹⁹

= 5²⁰¹⁷ . ( 1 + 5 - 5² )

= 5²⁰¹⁷ . ( -19) \(⋮\)19

=> 5²⁰¹⁷ + 5²⁰¹⁸ - 5²⁰¹⁹ \(⋮\)19

5+5^2+5^3+...+5^2006=(5+5^4)+(5^2+5^5)+(5^3+5^6)+...+(5^2003+5^2006).                                                                      =5.(1+5^3)+5^2.(1+5^3)+5^3.(1+5^3)+...+5^2003.(1+5^3).                                                               

= 5.126+5^2.126+5^3.125+...+5^2003.126

=126.(5+5^2+5^3+...+5^2003)chia hết cho 126. Vậy 5+5^2+5^3+...+5^2006 chia hết cho 126

\(\frac{3x-3}{6}=\frac{2y+10}{10}=\frac{5z-10}{15}=\frac{3x+2y-5z+17}{1}=\frac{3x+2y-5z+16+1}{1}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x-1}{2}=1\\\frac{y+5}{5}=1\\\frac{z-2}{3}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=0\\z=5\end{matrix}\right.\)

\(\Rightarrow P=3^{2019}+5^{2019}\)

Ta có \(3\equiv-1\left(mod4\right)\Rightarrow3^{2019}\equiv-1\left(mod4\right)\)

\(5\equiv1\left(mod4\right)\Rightarrow5^{2019}\equiv1\left(mod4\right)\)

\(\Rightarrow P\equiv\left(-1+1\right)\left(mod4\right)\Rightarrow P\equiv0\left(mod4\right)\Rightarrow P⋮4\)