Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(A=2+2^2+2^3+\dots+2^{60}\)
\(2A=2^2+2^3+2^4+\dots+2^{61}\)
\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)
\(A=2^{61}-2\)
Vậy: \(A=2^{61}-2\).
b)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\dots+\left(2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)
\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)
\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)
Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)
\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)
\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)
Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)
hay \(A\vdots5\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)
\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)
\(=7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)\)
Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)
$Toru$
a) �=2+22+23+⋯+260A=2+22+23+⋯+260
2�=22+23+24+⋯+2612A=22+23+24+⋯+261
2�−�=(22+23+24+⋯+261)−(2+22+23+⋯+260)2A−A=(22+23+24+⋯+261)−(2+22+23+⋯+260)
�=261−2A=261−2
Vậy: �=261−2A=261−2.
b)
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22)+(23+24)+(25+26)+⋯+(259+260)=(2+22)+(23+24)+(25+26)+⋯+(259+260)
=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)
=2⋅3+23⋅3+25⋅3+⋯+259⋅3=2⋅3+23⋅3+25⋅3+⋯+259⋅3
=3⋅(2+23+25+⋯+259)=3⋅(2+23+25+⋯+259)
Vì 3⋅(2+23+25+⋯+259)⋮33⋅(2+23+25+⋯+259)⋮3 nên �⋮3A⋮3
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)
=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)
=2⋅15+25⋅15+29⋅15+⋯+257⋅15=2⋅15+25⋅15+29⋅15+⋯+257⋅15
=15⋅(2+25+29+⋯+257)=15⋅(2+25+29+⋯+257)
Vì 15⋮515⋮5 nên 15⋅(2+25+29+⋯+257)⋮515⋅(2+25+29+⋯+257)⋮5
hay �⋮5A⋮5
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)
=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)
=2⋅7+24⋅7+27⋅7+⋯+258⋅7=2⋅7+24⋅7+27⋅7+⋯+258⋅7
=7⋅(2+24+27+⋯+258)=7⋅(2+24+27+⋯+258)
Vì 7⋅(2+24+27+⋯+258)⋮77⋅(2+24+27+⋯+258)⋮7 nên �⋮7A⋮7
a=2+2^2+2^3+...+2^10
a=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)
a=2.(1+2)+2^3.(1+2)+...+2^9.(1+2)
a=3.(2+2^3+...+2^9)
=> a chia hết cho 3
a=2+2^2+2^3+...+2^10
a=(2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)
a=2.(1+2+4+8+16)+2^6.(1+2+4+8+16)
a=31.(2+2^6)
=> a chia hết cho 31
chúc bạn học tốt nha
Sửa đề: \(A=2^0+2^1+2^2+...+2^{99}\)
\(=\left(2^0+2^1\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)
\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{98}\left(1+2\right)\)
\(=3\left(1+2^2+...+2^{98}\right)⋮3\)
\(A=2+2^2+...+2^{59}+2^{60}\)
\(A=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(A=2\cdot3+...+2^{59}\cdot3\)
\(A=3\cdot\left(2+...+2^{59}\right)⋮3\left(đpcm\right)\)
Bài 1:
a,\(A=3+3^2+3^3+...+3^{2010}\)
\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)
\(=3\left(1+3+3^2+3^3\right)+....+3^{2007}\left(1+3+3^2+3^3\right)\)
\(=3.40+...+3^{2007}.40\)
\(=40\left(3+3^5+...+3^{2007}\right)⋮40\)
Vì A chia hết cho 40 nên chữ số tận cùng của A là 0
b,\(A=3+3^2+3^3+...+3^{2010}\)
\(3A=3^2+3^3+...+3^{2011}\)
\(3A-A=\left(3^2+3^3+...+3^{2011}\right)-\left(3+3^2+3^3+...+3^{2010}\right)\)
\(2A=3^{2011}-3\)
\(2A+3=3^{2011}\)
Vậy 2A+3 là 1 lũy thừa của 3
A = 2 + 22 + 23 +.......+260
2A = 22 + 23+........+260+ 261
2A - A = 261 - 2
A = 261 - 2
A = 2 + 22 + 23+........+260
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 ⋮ 3 ⇔ A = 3.(2 + 23 +.....+259) ⋮ 3 (đpcm)
A = 261 - 2 = (24)15.2 - 2 = \(\overline{...6}\).2 - 2 = \(\overline{....2}\) - 2 = \(....0\) ⋮ 5
⇔ A ⋮ 5 (đpcm)
A = 2 + 22 + 23 +........+260
A = (2 +22+23)+(24+25+26)+.........+(258+259+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)
vì 7 ⋮ 7 ⇔ A = 7.(2+24+258) ⋮ 7 (đpcm)