\(a,S=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\\ S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ S=\left(3+3^2\right)\left(1+3^2+...+3^{18}\right)=12\left(1+3^2+...+3^{18}\right)⋮12\)

\(b,S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ S=\left(3+3^2+3^3+3^4\right)+....+3^{16}\left(3+3^2+3^3+3^4\right)\\ S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ S=120\left(1+...+3^{16}\right)⋮120\)

\(a,S=3+3^2+3^3+...+3^{20}\)

Ta thấy:\(3+3^2=12⋮12\)

\(\Rightarrow S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ \Rightarrow S=\left(3+3^2\right)\left(1+3^2+...+1^{18}\right)\\ \Rightarrow S=12.\left(1+3^2+...+3^{18}\right)⋮12\\ \left(đpcm\right)\)

\(b,Ta\) \(thấy:\)\(3+3^2+3^3+3^4=120⋮120\)

\(\Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+3^{16}\left(3+3^2+3^3+3^4\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ \Rightarrow S=120\left(1+...+3^{16}\right)⋮120\\ \left(đpcm\right)\)

(3+3²+3³)+(3⁴+3⁵+3⁶)+...+(3²⁰⁰⁵+3²⁰⁰⁶+3²⁰⁰⁷)

=3(1+3+3²)3⁴(1+3+3²)+...+3²⁰⁰⁵(1+3+3²)

=3.13+3^4.13+...+3^2005.13

=13(3+3⁴+...+3²⁰⁰⁵)

tick mk nha

Ta có 3.S=3.(3+3^2+3^3+........+3^2007)

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

Bài 1:x là số chẵn(x\(\in\)N)

bai 1 :x la so chan (chia het cho 2)

         x la so le (khong chia het cho 2

bai 2:tong cua 5 so tu nhien lien tiep chia het cho 5 vi tong 5 so tu nhien lien tiep la so co tan cung 0,5

bai 3:b,xy+yx=(x nhan 10)+y+(y nhan 10)+x=10x+y+10y+x=11x+11y.11x va 11y chia het cho 11. vay xy+yx chia het cho 11