a: \(G=8^8+2^{20}\)

\(=2^{24}+2^{20}\)

\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)

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

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

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

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

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

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

\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)⋮15\)

c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)

\(E=1+3+3^2+3^3+...+3^{1991}\)

\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)

\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)

l = 1 + 3 + 3² + ... + 3¹⁹⁹¹

l = (1 + 3 + 3²) + ... + (3¹⁹⁸⁹ + 3¹⁹⁹⁰ + 3¹⁹⁹¹)

l = 1.13 + ...+ 3¹⁹⁸⁹.13

l = 13.( 1 + .... + 3¹⁹⁸⁹)

=> I chia hết cho 13

a) \(B=3+3^2+3^3+...+3^{120}\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{199}\left(1+3\right)\)

\(=3.4+3^3.4+3^{199}.4=4\left(3+3^3+...+3^{199}\right)⋮4\)

b) \(B=3+3^2+3^3+...+3^{120}\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{198}\left(1+3+3^2\right)\)

\(=3.13+3^4.13+...+3^{198}.13=13\left(3+3^4+...+3^{198}\right)⋮13\)

A= 1 + 3 + 3^2 + 3^3 + 3^4 + ....+ 3^1991 

A= (1 + 3 + 3^2) +( 3^3 + 3^4+3^5) + ....+(3^1989+3^1999+3^1991)

A= 13+3^3(1+3+3^2)+....+3^1989(1+3+3^2) chia hết cho 13

Còn 41 thì gộp 4 số rùi làm tương tự

Chia hết cho 13 thì nhóm 3 số thành 1 cặp 

Ta có:

E=1+3+3^2+3^3+3^4+...+3^1991

E=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^1989+3^1990+3^1991)

E=13+3^3(1+3+3^2)+...+3^1989(1+3+3^2)

E=13+3^3.13+...+3^1989.13

E=13(1+3^3+...+3^1989) chia hết cho 13

còn chung minh chia hết cho 41 thì mik không biết

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

\(A=3.\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)...+3^{97}.\left(1+3+3^2\right)\)

\(A=3.13+3^4.13+...+3^{97}.13\)

\(A=13.\left(3+3^4+..+3^{97}\right)⋮13\)

Vậy...

\(A=3+3^2+3^3+...+3^{99}\)

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

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

\(A=3\cdot13+...+3^{97}\cdot13\)

\(A=13\cdot\left(3+...+3^{97}\right)⋮13\left(đpcm\right)\)

\(A=3+3^2+3^3+......+3^{99}\)

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

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

\(\Rightarrow A=3.13+.......+3^{97}.13\)

\(\Rightarrow A=13\left(3+....+3^{97}\right)⋮13\left(đpcm\right)\)

số số hạng của dãy là (1991-1) /2 +1=996 (số )

Vì 996 chia hết cho 3(bạn ghi kí hiệu vào )nên ta nhóm 3 số hạng liên tiêp

ta có ;3^1+3^3+3^5+3^7+......+3^1991

=(3^1+3^3+3^5)+(3^7+3^9+3^11)+.....+(3^1987+3^1989+3^1991)

=(3^1+3^3+3^5 *1)+(3^1 +3^3 +3^5 * 3^6)+....+(3^1+3^3+3^5 *3^1986)

=(3+27+243)+(3+27+ 243 * 3^6 )+...+(3+27+243 *3^1986)

=273+273 * 3^6+..... + 273* 3^1986

 =273 *(  1+ 3^6 +...+ 3^1986)

Vì 273  chia hết cho 13 Nên 273* (1+3^6+......+3^1986) chia hết cho 13

hay A chia hết cho 13

Vậy a chia hết cho 13

( bạn có thể thay nhung  chỗ VDchia hêt cho = kí hiêu đã học)

E = 1 + 3 + 3² + ....... + 3¹⁹⁹¹

E = ( 1 + 3 + 3² ) + ............. + ( 3¹⁹⁸⁹ + 3¹⁹⁹⁰ + 3¹⁹⁹¹ )

E = 1 . ( 1 + 3 + 3² ) + ............. + 3¹⁹⁸⁹ . ( 1 + 3 + 3² )

E = 1 . 13 + .............. + 3¹⁹⁸⁹ . 13

Mà 13 \(⋮\)13 nên E chia hết cho 13 ( đpcm )

Tương tự chia hết cho 41

A=3+3³+3⁵+...+3¹⁹⁹¹

A=(3+3³+3⁵)+...+(3¹⁹⁸⁷+3¹⁹⁸⁹+3¹⁹⁹¹)

A=3.(1+3²+3⁴)+...+3¹⁹⁸⁷.(1+3²+3⁴)

A=3.91+...+3¹⁹⁸⁷.91

A=3.7.13+...+3¹⁹⁸⁷.7.13

A=13.(3.7+...+3¹⁹⁸⁷.7) chia hết cho 13 (đcpm)