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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: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
\(E=1+3+3^2+3^3+...+3^{1991}\)
\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)
\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)
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
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...
\(A=3+3^2+3^3+...+3^{99}\)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
\(A=3\left(1+3+3^2\right)+...+3^{97}\left(1+3+3^2\right)\)
\(A=3\cdot13+...+3^{97}\cdot13\)
\(A=13\cdot\left(3+...+3^{97}\right)⋮13\left(đpcm\right)\)
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
