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a: \(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{48}\left(1+2\right)\)
\(=3\left(1+2^2+...+2^{48}\right)⋮3\)
b: \(2^0+2^1+2^2+...+2^{101}\)
\(=\left(1+2+2^2\right)+...+2^{99}\left(1+2+2^2\right)\)
\(=7\left(1+...+2^{99}\right)⋮7\)
c: 2A=2+2^2+...+2^101
=>A=2^101-1
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
\(M=2+2^2+...+2^{60}\)
\(=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\cdot\left(2+...+2^{59}\right)⋮3\)
\(M=2+2^2+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
a: \(2A=2^2+2^3+...+2^{61}\)
=>A=2^61-2
b: \(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{55}+2^{58}\right)\) chia hết cho 7(1)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)=3\left(2+2^3+...+2^{59}\right)⋮3\left(2\right)\)
Từ (1), (2) suy ra A chia hết cho 21
A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3
=>A chia hết cho 3
A= (2+22+23)+...+(258+259+260)
A=2.(1+2+22)+...+258.(1+2+22)
A=2.7+...+258.7
A=7.(2+...+258)
Vì 7 chia hết cho 7 =>7.(2+...+258) chia hết cho 7
CHIA HẾT CHO 3 :
A= (2+22)+(23+24)+...+(259+260)
A=2.(1+2)+23.(1+2)+...+259.(1+2)
A=2.3+23.3+...+259.3
A=3.(2+23+...+259)
Vì 3 chia hết cho 3 => 3.(2+23+...+259) chia hết cho 3
=>A chia hết cho 3
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: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
\(E=1+3+3^2+3^3+...+3^{1991}\)
\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)
\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
Lời giải:
$A=(2+2^2+2^3)+(2^4+2^5+2^6)+....+(2^{58}+2^{59}+2^{60})$
$=2(1+2+2^2)+2^4(1+2+2^2)+....+2^{58}(1+2+2^2)$
$=(1+2+2^2)(2+2^4+....+2^{58})$
$=7(2+2^4+....+2^{58})\vdots 7$.
A = 2+22+23+...+260
A = 2.(1+2+22) + 24.(1+2+22) + ... + 258.(1+2+22)
A = 2.7+24.7+...+258.7
A= 7. (2+24+...+258) chia hết cho 7
--> A chia hết cho 7 (ĐPCM)
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
