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\(C=\left(1+3+3^2+3^3\right)+...+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(C=\left(1+3+3^2+3^3\right)+...+3^8\left(1+3+3^2+3^3\right)\)

\(C=\left(1+3+3^2+3^3\right).\left(1+3^4+3^8\right)\)

\(C=40.\left(1+3^4+3^8\right)\)

Vậy \(C⋮40\)

sửa đề là cho \(C=1+3+3^2+3^3+...+3^{11}\)

Ta có: \(C=1+3+3^2+3^3+3^4+...+3^{11}\)

\(C=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+...+3^{10}\left(1+3\right)\)

\(C=4+3^2.4+3^4.4+3^6.4+...+3^{10}.4\)

\(C=4\left(1+3^2+3^4+3^6+3^8+3^{10}\right)⋮4\left(ĐPCM\right)\)

VẬy C chia hết cho 4

Ta có: C=(\(1+3+3^2+3^3\))+.......+\(\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(=40+.....+3^8\left(1+3+3^2+3^3\right)\)

\(=40\left(3^4+...+3^8\right)\)

Vậy \(C\)chia hết cho 40(Vì có chứa thừa số 40)

con khong biet

Sai hết :)

\(C=1+3+3^2+3^3+...+3^{11}\)

\(C=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{10}+3^{11}\right)\)

\(C=4+3^2.\left(1+3\right)+...+3^{10}.\left(1+3\right)\)

\(C=4+3^2.4+...+3^{10}.4\)

\(C=4.\left(1+3^2+...+3^{10}\right)\)

Vif \(4⋮4=>C⋮4\)

\(C=1+3+3^2+3^3+...+3^{11}\\ =\left(1+3\right)+3^2\left(1+3\right)+...+3^{10}\left(1+3\right)\\ =\left(1+3\right)\left(1+3^2+....+3^{10}\right)\\ =4\left(1+3^2+....+3^{10}\right)⋮4\)

\(=>C⋮4\)

C=1+3+32+33+...+311C=1+3+32+33+...+311

C=(1+3)+(32+33)+...+(310+311)C=(1+3)+(32+33)+...+(310+311)

C=4+32.(1+3)+...+310.(1+3)C=4+32.(1+3)+...+310.(1+3)

C=4+32.4+...+310.4C=4+32.4+...+310.4

C=4.(1+32+...+310)C=4.(1+32+...+310)

Vif 4⋮4=>C⋮4

(3¹ + 3² +3³ ) + (3⁴ + 3⁵ +3⁶ ) + ... + (3²⁰⁰⁸ + 3²⁰⁰⁹ + 3²⁰¹⁰ )

= 3 ( 1+ 3 + 9 ) + 3⁴ ( 1+ 3 +9 ) + ... + 3²⁰⁰⁸ ( 1 + 3 +9 )

= 13 ( 3 + 3⁴ + ... + 3²⁰⁰⁸ )    chia hết cho 13

Sửa đề: \(A=2^0+2^1+2^2+...+2^{99}\)

\(=\left(2^0+2^1\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)

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

\(=3\left(1+2^2+...+2^{98}\right)⋮3\)

Tôi  tên  là  Ngọc  Anh  . Năm  nay  Tôi 11 tuổi.  Tôi  không  biết  bài  này  

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