choA = 2 + 2^2 +2^3 +2^4+...+ 2^60
chung to rang A : 3
Cho A = 2+2^2+2^3+2^4+...+2^60. Chung to rang A chia het cho 3;7va 15
Ta có :
=2+2^2+2^3+...+2^60 = 2(1+2+2^2+2^3) + 2^5(1+2+2^2+2^3) + ... + 2^57(1+2+2^2+2^3)
A=(2+2^5+...+2^57)*15 chia het cho 15
CM:
A chia hết cho 21
=> A chia hết cho 3 và 7
Ta có
A=2(1+2)+2^3(1+2)+..............+2^59(1...
A=3(2+2^3+2^5+........+2^59)chia hết cho 3
Ta có :
A=2(1+2+2^2)+2^4(1+2+2^2)+...........+2...
A=7(2+2^4+2^7+..........+2^58)
=> A chia hết cho 3 và 7=> A chia hết
Vậy A chia hết cho 21 và 15
cho A = 2+ 2^2+2^3+2^4+....+2^60
chung to rang A chia het cho 3
\(A=2\left(2+1\right)+2^3\left(2+1\right)+2^5\left(1+2\right)+.....+2^{59}\left(2+1\right)\)
\(=2.3+2^3.3+2^5.3+.....+2^{59}.3\)
Vậy \(A⋮3\)
Chứng tỏ rằng: 1/3^2+1/4^2+1/5^2+...+1/60^2<4/9
kho vler ai biet thi tra loi gium!
cho A= 2+22+23+24+25+..........+260
chung to rang A chia het cho 3,7,15
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 chia hết cho 3 suy ra A chia hết cho3
A=(2+2^2+2^3)+(2^4+2^5+2^6)+.........+(2^58+2^59+2^60)=(2.1+2.2+2.2^2)+(2^4.1+2^4.2+2^4.2^2)+....+(2^58.1+2^58.2+2^58.2^2)
=2.(1+2+2^2)+2^4.(1+2+2^2)+.....+2^58.(1+2+2^2)
=2.7+2^4.7+...........+2^58.7 chia hết cho 7 suy ra A chia hết cho 7
câu A chia hết cho 15 bn gộp 4 số hạng lại với nhau nhé, nếu ko biết làm thì nhắn tin hỏi mk, mk giải ra cho
chung to rang:
A= 2 + 2^ + 2^3 + 2^4 + ... + 2^90
A= 2+2^2+2^3+2^4+...+2^90
=(2+2^90)+(2^2+2^89)+(2^3+2^88)+...
=2^91+2^91+2^91+...
=2^91.45
A=2+2^2+2^3+...+2^90
2.A=2^2+2^3+2^4+......+2^91
2.A-A=2^91-2
⇒A=2^91-2
Vậy A=2^91-2
chung to rang:
A= 2 + 2^2 + 2^3 + 2^4 + ...+ 2^90
chia het cho 21 bn Nguyễn Phúc Thịnh
chung minh rang
A= 2 mũ 1+ 2 mũ 2+ 2 mũ 3+ ....... +2 mũ 60
a] chung minh rang A chia hết cho 7
b] tìm n biết A+2 =2 mũ n
Cho A=2+2^2+2^3+...+2^60
Chung minh rang A chia het cho 6
A=2+2^2+2^3+...+2^60
A=(2+2^2)+(2^3+2^4)+...+(2^59+2^60)
A=6+2^2.(2+2^2)+...+2^58.(2+2^2)
A=6+2^2.6+...+2^58.6
A=6.(1+2^2+...+26^58)
Vì 6\(⋮\)6
=>6.(1+2^2+...+2^58) \(⋮\)6
=>A\(⋮\)6
Vậy A chia hết cho 6
B=\(^{2+2^2+2^3+...+2^{59}+2^{60}}\)
a, chung to rang B chia het cho 3,5,7
b,tim chu so tan cung cua B
B = 2+22+23+....+259+260
B = (2+22+23+24) +....+ (257+258+559+560)
B = 2(1+2+22+23)+...+ 257(1+2+22+23)
B = 2x15 +....+ 257x15
B = 15( 2+....+257) =>chia hết cho 5 vì 15 chia hết cho 5
a) B=2+22 + 23 + ...+ 259 + 260
B= (2+22) + (23+24) + .... + ( 259+ 260)
B= 2(1+2) + 23(1+2) + ... +259(1+2)
B= 2x3 + 23x3 + ... + 259x3
B= 3(2+23+......+259) => chia hết cho 3
B = 2+22+23+...+259+260
B = (2+22+23)+....+( 258+259+260)
B = 2(1+2+22) +...+ 258(1+2+22)
B = 2x7 +...+ 258x7
B = 7(2+...+258) => chia hết cho 7