S=(5+5²)+(5³+5⁴)+....+(5²⁰¹⁷+5²⁰¹⁸)

= 30+5²(5+5²)+....+5²⁰¹⁶(5+5²)

=30+30.5²+....+30.5²⁰¹⁶

^{vì từng số hạng của S chia hết cho 30 nên S chia hết cho 30}

a) \(S=5+5^2+5^3+5^4+...+5^{99}\)

\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{97}+5^{98}+5^{99}\right)\)

\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{97}\left(1+5+5^2\right)\)

\(=5.31+5^4.31+...+5^{97}.31\)

\(=31\left(5+5^4+...+5^{97}\right)⋮31\left(đpcm\right)\)

b) \(S=5+5^2+5^3+5^4+...+5^{99}\)

\(=5+\left(5^2+5^3\right)+\left(5^4+5^5\right)+...+\left(5^{98}+5^{99}\right)\)

\(=5+5\left(5+5^2\right)+5^3\left(5+5^2\right)+...+5^{97}\left(5+5^2\right)\)

\(=5+5.30+5^3.30+...+5^{97}.30\)

\(=5+30.\left(5+5^3+...+5^{97}\right)\)

Mà \(5⋮̸30\) nên \(S⋮̸30\left(đpcm\right)\)

c) Ta có: \(5S=5^2+5^3+5^4+5^5+...+5^{100}\)

\(5S-S=\left(5^2+5^3+5^4+5^5+...+5^{100}\right)-\left(5+5^2+5^3+5^4+...+5^{99}\right)\)

\(4S=5^{100}-5\)

\(\Rightarrow25^x-5=5^{100}-5\)

\(\Rightarrow25^x=5^{100}\)

\(\Rightarrow25^x=25^{50}\)

\(\Rightarrow x=50\)

S = 5 + 5² + 5³ + 5⁴ + .......... + 5⁹⁹

a)  S = ( 5 + 5² + 5³ ) + ( 5⁴ + 5⁵ + 5⁶ ) + .... + ( 5⁹⁷ + 5⁹⁸ + 5⁹⁹ )

    = 5. ( 1 + 5 + 5² ) + 5⁴ . ( 1 + 5 + 5² ) + .... + 5⁹⁷ . ( 1 + 5 + 5² )

     = ( 1 + 5 + 5² ). ( 5 + 5⁴ + .. + 5⁹⁷ )

      = 31 . ( 5 + 5⁴ + .... + 5⁹⁷ ) chia hết cho 31 ( đpcm )

c ) 5S = 5² + 5³ + .. + 5¹⁰⁰

=> 5S - S = 4S = 5¹⁰⁰ + 5⁹⁹ + ........ + 5³ + 5² - 5 - 5² - 5³ - ..... - 5⁹⁹

                         = 5¹⁰⁰ - 5 

25^x - 5 = 4S

=> 25^x - 5 = 5¹⁰⁰ - 5

=> 25^x = 5¹⁰⁰

=> 25^x = ( 5² )⁵⁰

=> 25^x = 25⁵⁰

=> x = 50

Vậy  x = 50

Câu b quên cách làm rồi     

a) S=5+52+53+54+...+599

=(5+52+53)+(54+55+56)+...+(597+598+599)

=5(1+5+52)+54(1+5+52)+...+597(1+5+52)

=5.31+54.31+...+597.31

=31(5+54+...+597)⋮31(đpcm)

b) S=5+52+53+54+...+599

=5+(52+53)+(54+55)+...+(598+599)

=5+5(5+52)+53(5+52)+...+597(5+52)

=5+5.30+53.30+...+597.30

=5+30.(5+53+...+597)

Mà 5⋮̸30 nên S⋮̸30(đpcm)

c) Ta có: 5S=52+53+54+55+...+5100

5S−S=(52+53+54+55+...+5100)−(5+52+53+54+...+599)

4S=5100−5

⇒25x−5=5100−5

⇒25x=5100

⇒25x=2550

⇒x=50

có cái báo cáo rồi

\(S=5+5^2+5^3+5^4+...+5^{2022}\)

\(S=\left(5+5^2\right)+\left(5^3+5^4\right)+\left(5^5+5^6\right)+...+\left(5^{2021}+5^{2022}\right)\)

\(S=\left(5+5^2\right)+5^2\cdot\left(5+5^2\right)+5^4\cdot\left(5+5^2\right)+...+5^{2020}\cdot\left(5+5^2\right)\)

\(S=\left(5+5^2\right)\left(1+5^2+5^4+...+5^{2020}\right)\)

\(S=30\left(1+5^2+5^4+...+5^{2020}\right)\)

Vậy S chia hết cho 30

S không thể chia hết cho 13 nhé

S=(5+5^2)+5^2(5+5^2)+...+5^2020(5+5^2)

=30*(1+5^2+...+5^2020) chia hết cho 30

Cho 2 dễ rồi bạn tự làm nhé 
Cho 5 
2 + 2^2 + .... + 2^ 30 
(2 + 2^3 ) + ( 2^ 2 + 2^ 4) + ..... = ( 2^28 + 2^30 ) 
2( 1 + 4 ) + 2^2 ( 1 + 4 ) +...... + 2^28 ( 1+ 4 ) 
2. 5 + ..... + 2^ 28 . 5 
5( 2 + >.. + 2^ 28 ) chia hết cho 5

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

1/5 S = 1+5+5^2+...+5^2012

         =1(1+5+5^2)+5^3(1+5+5^2)+...+5^2010(1+5+5^2)

        mà 1+5+5^2=31=>1+5+5^2 chia hết 31

        => mổi số hạng của 1/5 S chia hết 31

       => S chia hết 31

Học chuyên đó ak. bài zễ thế nài mà ko bt làm ntn hả

ta có : S=5+5^2+5^3+5^4+......+5^2013  ( có 2013 số hạng )

           S=(5+5^2+5^3)+(5^4+5^5+5^6)+.............+(5^2011+5^2012+5^2013)   ( có 671 nhóm)

           S= 5.(1+5+5^2)+5^2.(1+5+5^2)+........+5^2011.(1+5+5^2)

           S=(5+5^2+.....+5^2011).31

            S chia hết cho 31                

S = (5 + 5²) + (5³ + 5⁴) +....+(5⁹ + 5¹⁰)

S = 1.30 + 5².30+....+5⁸.30

S = 30.(1+5²+....+5⁸)

S chia hết cho 30

=> ĐPCM 

\(S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^9+5^{10}\right)\)

\(=30+5^2.\left(5+5^2\right)+...+5^8.\left(5+5^2\right)\)

\(=30+5^2.30+...+5^8.30\)

\(=30.\left(1+5^2+...+5^8\right)\text{ chia hết cho 30}\)

=> S chia hết cho 30 (đpcm).