thieu de hay sao y

cuối là chia hết cho 6

Đặt A=5+5²+5³+5⁴+...+5³⁰

=(5+5²)+(5³+5⁴)+...+(5²⁹+5³⁰)

=5.(1+5)+5³(1+5)+...+5²⁹(1+5)

=5.6+5³.6+...+5²⁹.6

Vì 6 chia hết cho 6 nên 5.6+5³.6+...+5²⁹.6 chia hết cho 6

hay A chia hết cho 6

Vậy A chia hết cho 6

\(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+.............+3^{24}\left(3+2^3+3^5\right)\)

\(B=273+273\cdot3^6+.............+273\cdot3^{24}\)

\(B=273\left(1+3^6+.......+3^{24}\right)⋮273\)

\(A=\left(5+5^2\right)+\left(5+5^2\right)5^2+\left(5+5^2\right)5^4+\left(5+5^2\right)5^6+\left(5+5^2\right)5^8\)

\(A=30+30\cdot5^2+30\cdot5^4+30\cdot5^6+30\cdot5^8\)

\(A=30\left(1+5^2+5^4+5^6+5^8\right)⋮30\)

cảm ơn 

Đặt : \(A=5+5^2+5^3+...+5^{30}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{29}\left(1+5\right)\)

\(=\left(1+5\right)\left(5+5^3+...+5^{29}\right)\)

\(=6\left(5+5^3+...+5^{29}\right)⋮6\) (đpcm)

                                                   Bài giải

\(5+5^2+5^3+5^4+...+5^{29}+5^{30}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{29}+5^{30}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{29}\left(1+5\right)\)

\(=5\cdot6+5^3\cdot6+...+5^{29}\cdot6\)

\(=6\left(5+5^3+...+5^{29}\right)\text{ }⋮\text{ }6\)

\(\Rightarrow\text{ ĐPCM}\)

  S= 5+5²+5³+...+5²⁰²⁰+5²⁰²¹

 5S=5²+5³+5⁴+...+5²⁰²¹+5²⁰²²

 5S - S=4S=5²⁰²²-5

  Ta có: 4S+5=5²⁰²²

             =4S -5 +5 =5²⁰²²

              => 4S=5²⁰²²

a) \(A=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^6\left(5+5^2\right)=30+5^2.30+...+5^6.30\)

\(=30\left(1+5^2+...+5^6\right)⋮30\Rightarrowđpcm\)

b) \(B=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^3+3^5\right)=273+3^6.273+...+3^{24}.273\)

\(=273.\left(1+3^6+...+3^{24}\right)⋮273\Rightarrowđpcm\)

a: \(B=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)\)

\(=156\cdot5\cdot\left(1+5^4\right)\)

\(=780\left(1+5^4\right)⋮30\)

b: \(B=\left(3+3^3+3^5\right)+...+3^{24}\left(3+3^2+3^5\right)\)

\(=273\cdot\left(1+...+3^{24}\right)⋮273\)

ai trả lời giúp mình

1^3+2^3+4^3+5^3=(1+2+3+4+5)^2

1³ + 2³ + 3³ + 4³ + 5³ = 1 + 8 + 27 + 64 + 125 

                                    = 225

( 1 + 2 + 3 + 4 + 5 )² = 15² = 225

\(\Rightarrow\)1³ + 2³ + 3³ + 4³ + 5³ = ( 1 + 2 + 3 + 4 + 5 )²

Hk tốt

\(S=2^1+2^2+2^3+2^4+2^5+2^6+..+2^{28}+2^{29}+2^{30}\) 

\(S=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{28}.\left(1+2+2^2\right)\) 

\(S=\left(1+2+2^2\right).\left(2+2^4+...+2^{28}\right)\) 

\(S=7.\left(2+2^4+...+2^{28}\right)\) 

⇒ \(S⋮7\)   ( điều phải chứng minh ) 

S=2¹+2²+2³+...+2³⁰

S=(2¹+2²+2³)+(2⁴+2⁵+2⁶)+...+(2²⁸+2²⁹+2³⁰)

S=7.2+7.2⁴+...+7.2²⁸

S=7.(2+2⁴+...+2²⁸)

⇒S⋮7

Ta có: \(S=2^1+2^2+2^3+...+2^{28}+2^{29}+2^{30}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{28}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+2^4+...+2^{28}\right)⋮7\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); ...; \(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

=> S < \(5\left(1-\frac{1}{100}\right)=5.\frac{99}{100}< 5.1=5\)=> S<5

Lại có: \(\frac{1}{2^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\); \(\frac{1}{3^2}>\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\); \(\frac{1}{100^2}>\frac{1}{100.101}=\frac{1}{100}-\frac{1}{101}\)

=> \(S>5\left(\frac{1}{2}-\frac{1}{101}\right)=5.\frac{101-2}{2.101}=\frac{5.99}{2.101}~2,45\)=> S>2

Vậy 2 < S < 5 => Đpcm