\(C=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)...+\left(5^{17}+5^{18}+5^{19}+5^{20}\right)\\ C=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)...+5^{17}\left(1+5+5^2+5^3\right)\\ C=5\cdot156+5^5\cdot156+...+5^{17}\cdot156\\ C=156\left(5+5^5+...+5^{17}\right)\\ C=12\cdot13\left(5+5^5+...+5^{17}\right)⋮17\)

(5 +5³)+(5²+5⁴)...+(5¹⁸+5²⁰)

5(1+5²)+5²(1+5²)+...+5¹⁸(1+5²)

(1+5²)(5+5²+...+5¹⁸)

26(5+5²+...+5¹⁸)⋮13

vậy (5 +5³)+(5²+5⁴)...+(5¹⁸+5²⁰)⋮13

 

ai giúp mik với rồi mik tik cho

 

\(A=5+5^2+5^3+...+5^{20}\)

\(A=\left(5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{18}+5^{19}+5^{20}\right)\)

\(A=30+5^3\cdot31+...+5^{18}\cdot31\)

\(A=30+31\cdot\left(5^3+5^6+...+5^{18}\right)\)

Mà: \(31\cdot\left(5^3+5^6+...+5^{18}\right)\) ⋮ 31

\(\Rightarrow A=30+31\cdot\left(5^3+5^6+...+5^{18}\right)\) chia cho 31 dư 30 

A = 5 + 5² + 5³ +...+ 5²⁰

A = 5²⁰ + 5¹⁹ + 5¹⁸ +...+ 5³ + 5² + 5

A = (5²⁰ + 5¹⁹ + 5¹⁸) + (5¹⁷ + 5¹⁶ + 5¹⁵) +...+ (5⁵ + 5⁴ + 5³) + (5²+ 5)

A = 5¹⁸.( 5² + 5 + 1) + 5¹⁵.(5² + 5 + 1) +...+ 5³.(5²+ 5 + 1) + (25 + 5)

A = 5¹⁸. 31 + 5¹⁵.31 +...+ 5³.31 + 30

A = 31.(5¹⁸ + 5¹⁵ +...+ 5³) + 30

31 ⋮ 31 ⇒ 31.(5¹⁸ + 5¹⁵ +...+5³) ⋮ 31 mà 30 : 31 = 0 dư 31 

Vậy A : 31 dư 30 

 

 

Ta có: \(\frac{1}{50}\) >\(\frac{1}{100}\)

\(\frac{1}{51}\)>\(\frac{1}{100}\)

\(\frac{1}{52}\)>\(\frac{1}{100}\)

..................

\(\frac{1}{99}\)>\(\frac{1}{100}\)

=>\(\frac{1}{50}\)+\(\frac{1}{51}\)+.............+\(\frac{1}{99}\)>\(\frac{1}{100}\).50=\(\frac{1}{2}\)(50 là số số hạng  của S nha)

=>S>\(\frac{1}{2}\)

   

A = 5 + 5² + 5³ +...+ 5⁸

A = (5 + 5²) +( 5³ + 5⁴) +...+ ( 5⁷ + 5⁸)

A = 30 + 5².(5 + 5²) +...+ 5⁶.(5 + 5²)

A = 30.( 1 + 5² +...+ 5⁶) (đpcm)

 

   

F = 7 + 72 + 73 + 74 + ..... + 7100 

F= 7+(1+7)+73+(1+7)+...+799+(1+7)

F = 7x8+73x8+...+799x8

F= 8x(7+73+...+799)

mà 8 chia hết 8 => 8(7+73+...+799) chia hết 8

Vậy F chia hết cho 8

2)

\(F=7+7^2+7^3+7^4+...+7^{100}\\ F=7\cdot\left(1+7\right)+7^3\cdot\left(1+7\right)+.....+7^{99}\cdot\left(1+7\right)\\F=7\cdot8+7^3\cdot8+.....+7^{99}\cdot8\\ F=8\cdot\left(7+7^3+....+7^{99}\right)\\ =>F⋮8\) 

Tick giúp mik với

 

\(a,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(5\left(1+5+5^2+\cdot\cdot\cdot+5^{19}\right)⋮5\)

nên \(C⋮5\)

\(b,C=5+5^2+5^3+5^4\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdot\cdot\cdot+\left(5^{19}+5^{20}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdot\cdot\cdot+5^{19}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+\cdot\cdot\cdot+5^{19}\cdot6\)

\(=6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)\)

Ta thấy: \(6\cdot\left(5+5^3+\cdot\cdot\cdot+5^{19}\right)⋮6\)

nên \(C⋮6\)

\(c,C=5+5^2+5^3+5^4+\cdot\cdot\cdot+5^{20}\)

\(=\left(5+5^3\right)+\left(5^2+5^4\right)+\cdot\cdot\cdot+\left(5^{17}+5^{19}\right)+\left(5^{18}+5^{20}\right)\)

\(=5\left(1+5^2\right)+5^2\left(1+5^2\right)+\cdot\cdot\cdot+5^{17}\cdot\left(1+5^2\right)+5^{18}\left(1+5^2\right)\)

\(=5\cdot26+5^2\cdot26+\cdot\cdot\cdot+5^{17}\cdot26+5^{18}\cdot26\)

\(=26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)\)

Ta thấy: \(26\cdot\left(5+5^2+\cdot\cdot\cdot+5^{17}+5^{18}\right)⋮13\)

nên \(C⋮13\)

#\(Toru\)

a, ta có
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 + ... + 5^19 )
=> C chia hết cho 5
b,
C = 5 + 5^2 + 5^3 + 5^4 + ... + 5^20
=> C = 5 . ( 1 + 5 ) + 5^3 . ( 1 + 5 ) + ... + 5^19 . ( 1 + 5 )
=> C = 5 . 6 + 5^3 . 6 + ... + 5^19 . 6
=> C = 6 . ( 5 + 5^3 + ... + 5^19 )
=> C chia hết cho 6
c,
C = 5 + 5^2 + 5^3 + ... + 5^20
=> C = (5 + 5^2 + 5^3 + 5^4 ) + ... + (5^17 + 5^18 + 5^19 + 5^20 )
=> C = 5 . ( 1 + 5 + 5^2 + 5^3 ) + ... + 5^17 . ( 1+ 5 + 5^2 +5^3)
=> C = 5 . 156 + 5^5 . 156 + ...+ 5^17 . 156
=> C = 5 . 12 . 13 + 5^5 . 12 . 13 + ... + 5^17 . 12 . 13
=> C = 13 . ( 5 . 12 + 5^5 . 12 + ... + 5^17 . 12 )
=> C chia hết cho 13