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).

S = (1+5)+(5^2+5^3)+(5^4+5^5)+(5^6+5^7)

   = 6+5^2.(1+5)+5^4.(1+5)+5^6.(1+5)

   = 6+5^2.6+5^4.6+5^6.6

   = 6.(1+5^2+5^4+5^6) chia hết cho 6

=> ĐPCM

k mk nha

(1+5)+(5^2+5^3)+........+(5^6+5^7)

=6+5^2(1+5)+......+5^6(1+5)

=6+5^2 . 6 +.....+5^6 . 6

= 6 ( 5^2+.....+5^6)

Suy ra S chia hết cho 6

\(S=\left(1+5\right)+\left(5^2+5^3\right)+\left(5^4+5^5\right)+\left(5^6+5^7\right)\)

\(=6+5^2\times6+5^4\times6+5^6\times6\)

\(=6\left(1+5^2+5^4+5^6\right)\)chia hết cho 6

=> S chia hết cho 6 =>ĐPCM

N=(5+5^2)+5^2x(5+5^2)+5^4x(5+5^2)+5^6x(5+5^2)+5^8x(5+5^2)

N=30+5^2x30+5^4x30+5^6x30+5^8x30

N=(1+5^2+5^4+5^6+5^8)x30

Vì 30 chia hết cho 6 nên N chia hết cho 6.

N=(5+5^2)+5^2x(5+5^2)+5^4x(5+5^2)+5^6x(5+5^2)+5^8x(5+5^2)

N=30+5^2x30+5^4x30+5^6x30+5^8x30

N=(1+5^2+5^4+5^6+5^8)x30

Vì 30 chia hết cho 6 nên N chia hết cho 6.

(1+5^2+5^4+5^6+5^8).x 5+5^3+5^5+ ... 5^9

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

Bài 3:

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

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

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

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

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

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

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

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}