\(=5^{20}+\left(5^2\right)^{11}+\left(5^{ }^3\right)^7\)

=\(5^{^{ }20}+5^{22}+5^{21}\)

\(=5^{20}\cdot\left(1+5^2+5^1\right)\)

=\(5^{20}\cdot\left(1+25+5\right)\)

=\(5^{20}\cdot31\)

Vì 31 chia hết chó 31 nên

\(5^{20}+25^{^{ }11}+125^7\)chia hết cho 31

\(^{5^{20}+25^{11}+125^7}\)=\(1.5^{20}+25.25^{10}+\left(5^3\right)^7\)=\(1.5^{20}+25.\left(5^2\right)^{10}+5^{21}\)=\(1.5^{20}+25.5^{20}+5.5^{20}\)

=\(^{5^{20}.\left(1+25+5\right)}\)=\(5^{20}.31\)chia hết cho 31

Vậy \(5^{20}+25^{11}+125^7\)chia hết cho 31

5^20+25^11+125^7=5^20+(5^2)^11+(5^3)^7= 5^20+5^22+5^21=5^20(1+5^2+5)=5^20.31

Vậy 5^20+25^11+125^7 chia hết cho 31

ta có(^ là dấu mũ):

5^20+25^11+125^7=5^20+5^22+5^21

=5^20+5^20.5^2+5^21.5

=5^20.(1+5^2+5)=5^20.(1+25+5)=5^20.31 chia hết cho 31

Nếu sai chỗ nào thì nhắc mik nhé :)

\(5^{20}+25^{11}+125^7=5^{20}+5^{2^{11}}+5^{3^7}=5^{20}+5^{22}+5^{21}=5^{20}+5^{20}.5^2+5^{20}.5=5^{20}\left(5^2+5+1\right)=5^{20}.31\)Vì \(5^{20}.31⋮31\) nên \(\left(5^{20}+25^{11}+125^7\right)⋮31\)

Ra A= 5^11-5^3

Vì 5^11chia hết 125

     5^3 chia hết cho125

=> 5^11-5^3 chia hết cho125

A=(5^11-5^3)/4

M = 125⁷ - 625⁵ - 25⁹

M = ( 5³ )⁷ - ( 5⁴ )⁵ - ( 5² )⁹

M = 5²¹ - 5²⁰ - 5¹⁸

M = 5¹⁸ . ( 5³ - 5² - 1 )

M = 5¹⁸ . 99

M = 5¹⁸ . 9 . 11 \(⋮\)9

5^61 + 25^31 + 125^21

= 5^61 + 5^62 + 5^63

= 5^61 x (1+5+25) 

= 5^61 x 31 chia hết 31

5^61 + 25^31 + 125^21

= 5^61 + 5^62 + 5^63

= 5^61 x (1+5+25) 

= 5^61 x 31 chia hết 31

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$