1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)\)

\(=30\left(1+2^4+...+2^{96}\right)⋮30\)

2:

\(B=3+3^2+3^3+...+3^{2022}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)

\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)

Ta có: A= 2 + 2² + 2³ + ... + 2⁶⁰= (2 +2²) + (2³+ 2⁴) + ... + (2⁵⁹ + 2⁶⁰).

             = 2 x (2 + 1) + 2³ x (2 + 1) + ... + 2⁵⁹ x (2 + 1).

             = 2 x 3 + 2³ x 3 + ... + 2⁵⁹ x 3.

             = 3 x ( 2 + 2³ + ... + 2⁵⁹).

Vì A = 3 x ( 2 + 2³ + ... + 2⁵⁹)  nên A chia hết cho 3.

           A= (2 +2² + 2³) + (2⁴ + 2⁵  + 2⁶) + ... + (2⁵⁸ + 2⁵⁹ + 2⁶⁰).

             = 2 x (1 + 2 + 2²) + 2⁴ x (1 + 2 + 2²) + ... + 2⁵⁸ x (1 + 2 + 2²).

             = 2 x 7 + 2⁴ x 7 + ... + 2⁵⁸ x 7.

             = 7 x ( 2 + 2⁴ + ... + 2⁵⁸).

Vì A = 7 x ( 2 + 2⁴ + ... + 2⁵⁸)  nên A chia hết cho 7.

  A= (2 +2² + 2³ + 2⁴) + (2⁵ + 2⁶  + 2⁷ + 2⁸) + ... + (2⁵⁷ + 2⁵⁸ + 2⁵⁹ + 2⁶⁰).

             = 2 x (1 + 2 + 2² + 2³) + 2⁵ x (1 + 2 + 2² + 2³) + ... + 2⁵⁷ x (1 + 2 + 2² + 2³).

             = 2 x 15 + 2⁵ x 15 + ... + 2⁵⁷ x 15.

             = 15 x ( 2 + 2⁴ + ... + 2⁵⁸).

Vì A = 15 x ( 2 + 2⁴ + ... + 2⁵⁸)  nên A chia hết cho 15.

Ta có: B= 3 + 3³ + 3⁵ + ... + 3¹⁹⁹¹= (3 + 3³ + 3⁵) + (3⁷+ 3⁹ + 3¹¹ ) + ... + (3¹⁹⁸⁷ + 3¹⁹⁸⁹ + 3¹⁹⁹¹).

             = 3 x (1 + 3² + 3⁴) + 3⁷ x (1 + 3² + 3⁴) + ... + 3¹⁹⁸⁷ x (1 + 3² + 3⁴).

             = 3 x 91 + 3⁷ x 91 + ... + 3¹⁹⁸⁷ x 91= 3 x 7 x 13 + 3⁷  x 7 x 13 + ... + 3¹⁹⁸⁷ x 7 x 13.

             = 13 x ( 3 x 7 + 3⁷ x 7 + ... + 3¹⁹⁸⁷ x 7).

Vì B = 13 x ( 3 x 7 + 3⁷ x 7 + ... + 3¹⁹⁸⁷ x 7) nên B chia hết cho 13.

           B= (3 + 3³ + 3⁵ + 3⁷) +  ... + (3¹⁹⁸⁵ + 3¹⁹⁸⁷ + 3¹⁹⁸⁹ + 3¹⁹⁹¹).

             = 3 x (1 + 3² + 3⁴  + 3⁶) +  ... + 3¹⁹⁸⁵ x (1 + 3² + 3⁴ ​+ 3⁶).

             = 3 x 820 + ... + 3¹⁹⁸⁵ x 820= 3 x 20 x 41 + ... + 3¹⁹⁸⁵ x 20 x 41.

             = 41 x ( 3 x 20 + .. +  3¹⁹⁸⁵ x 20)

Vì B =41 x ( 3 x 20 + .. +  3¹⁹⁸⁵ x 20) nên B chia hết cho 41.

a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)

\(=3\times91+3^7\times91+...+3^{1987}\times91\)

\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)

\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)

Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.

b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)

\(=3\times820+...+3^{1985}\times820\)

\(=3\times20\times41+...+3^{1985}\times20\times41\)

\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)

Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.

A={2+2^2}+{2^3+2^4}+.......+{2^59+2^60}

={2.1+2.2}+{2^3.1+2^3.2}+....+{2^59.1+2^59.2}

=2{1+2}+2^3{1+2}+...+2^59{1+2}

=2.3+2^3.3+.....+2^59.3

=3.(2+2^3+...+2^59)

vi co thua so 3 => tich do chia het cho 3

A={2+2^2}+{2^3+2^4}+.......+{2^59+2^60}

={2.1+2.2}+{2^3.1+2^3.2}+....+{2^59.1+2^59.2}

=2{1+2}+2^3{1+2}+...+2^59{1+2}

=2.3+2^3.3+.....+2^59.3

=3.(2+2^3+...+2^59)

vi co thua so 3 => tich do chia het cho 3

\(M=2+2^2+...+2^{60}\)

\(=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\cdot\left(2+...+2^{59}\right)⋮3\)

\(M=2+2^2+...+2^{60}\)

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

ê

giúp mình với mình chuẩn bị phải nộp bài rồi T~T 

\(B=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

a) A chia hết cho 2 vì tất cả các số hạng của tổng đều chia hết cho 2.

b) Ta tách ghép các số hạng của A thành các nhóm sao cho mỗi nhóm xuất hiện thừa số chia hết cho 3. Khi đó:

a) \(A=2+2^2+2^3+...+2^{20}\)

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

\(A=2\cdot\left(1+3\right)+2^3\cdot\left(1+3\right)+...+2^{59}\cdot\left(1+3\right)\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\)

Vậy A chia hết cho 3

________

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

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy A chia hết cho 5 

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=14+2^3\cdot14+...+2^{117}\cdot14\)

\(=14\cdot\left(1+2^3+...+2^{117}\right)⋮7\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=62+2^5\cdot62+...+2^{115}\cdot62\)

\(=62\cdot\left(1+2^5+...+2^{115}\right)⋮31\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=126+126\cdot2^6+...+126\cdot2^{114}\)

\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)

cho a+b+c=0 cmr

a^3 + b^3+a^2c+b^2c-abc=0

A=2+2²+2³+...+2⁶⁰

A=(2+2²+2³)+...+(2⁵⁸+2⁵⁹+2⁶⁰)

A=12.1+...+2⁵⁷.(2+2²+2³)

A=12.1+...+2⁵⁷.12

A=12.(1+...+2⁵⁷)chia hết cho  3 vì 12 chia hết cho 3

tương tự chia lần lượt thành 4 nhóm ,5 nhóm :b)thì chia lần lượt thành 3 nhóm,4 nhóm