S = 3¹⁰⁰ - 1

Ad cho xin ý kiến vs ạ

a)B=1+3+3²+3³+....+3⁹⁹

=(1+3)+(3²+3³)+...+(3⁹⁸+3⁹⁹)

=4+3².4+....+3⁹⁸.4

=4.(1+3²+...+3⁹⁸) chia hết cho 4

Vậy B chia hết cho 4

b)B=1+3²+3³+3⁴+...+3⁹⁹

=(1+3+3²+3³)+....+(3⁹⁶+3⁹⁷+3⁹⁸+3⁹⁹)

=40+.........+3⁹⁶.40

=40.(1+....+3⁹⁶) chia hết cho 40

Vậy B chia hết cho 40

a)B=(1+3)+(3²+3³)+...+(3⁹⁸+3⁹⁹)

=(1+3)+3²(1+3)+....+3⁹⁸(1+3)

=4+3².4+...+3⁹⁸.4

=4(1+3²+...+3⁹⁸) chia hết cho4

câu b bạn vận dụng theo câu a là đc bạn nhóm 4 lại nhé mình hơi lười làm

a)     B=3^0+3^1+3^2+  .............+3^99

=1(1+3)+3^2(1+3)+.................3^98(1+3)

=4+3^2.\(\times4+.............+3^{98}\times4\)

\(=4\left(1+3^2+............3^{98}\right)\)

\(\Rightarrow\)Bchia hết cho 4

C/M C\(⋮\)4

\(C=1+3+3^2+...+3^{99}⋮4\)

\(C=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)⋮4\)

\(C=\left(1+3\right)+3^2.\left(1+3\right)+...+3^{98}.\left(1+3\right)⋮4\)

\(C=4+3^2.4+...+3^{98}.4⋮4\)

\(C=4.\left(1+3^2+...+3^{98}\right)⋮4\)

C/M C\(⋮\)40

\(C=1+3+3^2+...+3^{99}⋮40\)

\(C=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)⋮40\)

\(C=\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)⋮40\)

\(C=40.1+...+3^{96}.40⋮40\)

\(C=40.\left(1+...+3^{96}\right)⋮40\)

ai lạnh ko tui lạnh quá mà vẫn ko có ng iu

 ĐẶt A=3+3^2+3^3+....+3^100 
Chia A thành từng nhóm 4 số (vì A có 100 số) ta được 25 nhóm 
A= 3(1+3+3^2+3^3) +3^5(1+3+3^2+3^3)+...... 
+3^97(1+3+3^2+3^3) 
A=3.40 +3^5.40+.....+3^97.40 
Vậy A chia hết cho 40. 

C=\(3+3^2+3^3+3^4+...+3^{100}\)

= \(3\times\left(1+3+3^2+3^3+...+3^{99}\right)\)

= \(3\times\left[\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\right]\)

= \(3\times\left(40+3^4\times40+...+3^{96}\times40\right)\)

= \(3\times40\times\left(1+3^4+...+3^{96}\right)\)chia het cho 40

=> C chia het cho 40

A=3+3²+.........+3²⁰¹⁶

A=3.(1+3+9+27)+.....+3²⁰¹³.(1+3+9+27)

A=3.40+.....+3²⁰¹³.40

A=40.(3+...+3²⁰¹³) 

=> A\(⋮40\)

=> ĐPCM .

\(\text{Ta có:}\)

\(B=1+3+3^2+3^3+3^4+3^5+3^6+3^7+.......+3^{96}+3^{97}+3^{98}+3^{99}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+.....+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

\(=40+\left[3^4\left(1+3+3^2+3^3\right)\right]+.....+\left[3^{96}\left(1+3+3^2+3^3\right)\right]\)

\(=40+3^4\cdot40+....+3^{96}\cdot40\)

\(=40\left(1+3^4+....+3^{96}\right)\)

\(\Rightarrow B⋮40\)

\(3+3^2+3^3+3^4+...+3^{100}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1\left(3+3^2+3^3+3^4\right)+...+3^{96}\left(3+3^2+3^3+3^4\right)\)

\(=\left(1+...+3^{96}\right)\left(3+3^2+3^3+3^4\right)\)

\(=Q.120\)

\(=Q.3.40\)

\(\Rightarrow3+3^2+3^3+3^4+...+3^{100}⋮40\) (Đpcm)