Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B = ( 1 + 2 ) + ( 2^2 + 2^3 ) + ( 2^4 + 2^5 ) + ... + ( 2^99 + 2^100 )
B = 1( 1 + 2 ) + 2^2( 1 + 2 ) + 2^4( 1 + 2 ) + ... + 2^99( 1 + 2)
B = 1 . 3 + 2^2 . 3 + 2^4 . 3 + ... + 2^99 . 3
B = 3( 1 + 2^2 + 2^4 + ... + 2^99 )
=> B chia hết cho 3
B = ( 1 + 2 + 2^2 ) + ( 2^3 + 2^4 + 2^5 ) + ... + ( 2^98 + 2^99 + 2^100 )
B = 1( 1 + 2 + 2^2 ) + 2^3( 1 + 2 + 2^2 ) + ... + 2^98( 1 + 2 + 2^2 )
B = 1.7 + 2^3.7 + ... + 2^98.7
B = 7( 1 + 2^3 + ... + 2^98 )
=> B chia hết cho 7
B = ( 1 + 2 + 2^2 + 2^3 ) + ( 2^4 + 2^5 + 2^6 + 2^7 ) + ... ( 2^97 + 2^98 + 2^99 + 2^100 )
B = 1( 1 + 2 + 2^2 + 2^3 ) + 2^4( 1 + 2 + 2^2 + 2^3 ) + ... + 2^97( 1 + 2 + 2^2 + 2^3 )
B = 1.15 + 2^4.15 + ... + 2^98.15
B = 15( 1 + 2^4 + ... + 2^98 )
=> B chia hết cho 15
Mà 15 = 3 . 5
=> B chia hết cho 5
a)\(B=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{99}+2^{100}\right)\)
\(B=\left(1+2\right)+2^2\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(B=3+2^2.3+...+2^{99}.3\)
\(B=3\left(1+2^2+...+2^{99}\right)⋮3\left(đpcm\right)\)
b)\(B=\left(1+2+2^2\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(B=\left(1+2+2^2\right)+...+2^{98}\left(1+2+2^2\right)\)
\(B=7+....+2^{98}.7\)
\(B=7\left(1+...+2^{98}\right)⋮7\left(đpcm\right)\)
c)\(B=\left(1+2+2^2+2^3\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(B=\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(B=15+...+2^{97}.15\)
\(B=15\left(1+...+2^{97}\right)⋮5\left(đpcm\right)\)
A = 31 + 32 +33 + 34 +.....+32015+ 32016
A = (31 + 32) +(33 + 34) +.....+ (32015+ 32016)
A = 3(1+3) + 32(1+3) + .....+ 32015(1+3)
A = 3.4 +32.4 +....... + 32015.4
A = 4(3 +32 +....+ 32015) chia hết cho 4
===================================================
A =31 + 32 +33 + 34 + 35 +36 +.....+32014 + 32015+ 32016
A = (31 + 32 +33 ) +(34 + 35 +36) +.....+ (32014 + 32015+ 32016)
A = 3(1+3+32) + 34(1+3+32) + .....+ 32014(1+3+32)
A = 3.13 +34.13 +....... + 32014.13
A = 13.(3 +34 +....+ 32014) chia hết cho 13
A = 3 + 32 + 33 + 34 +..... + 32015 + 32016
= (3 + 32 + 33) + (34+ 35 + 36 ) +.....+ (32014 + 32015 + 32016)
= 3(1 + 3 + 32) + 34(1 + 3 + 32) + .....+ 32014(1 + 3 + 32)
= 13(3 + 34 + ....+ 32014) \(⋮13\)
A = 3 + 32 + 33 + 34 +..... + 32015 + 32016
= (3 + 32) + (33 + 34) + .... + (32015 + 32016)
= 3(1 + 3) + 33(1 + 3) + .... + 32015(1 + 3)
= 4(3 + 33 + .... + 32015) \(⋮4\)
8 Cho A = 32016 + 32015 + ... + 32 + 3
a) Chứng minh A chia hết cho 4
b) Chứng minh A chia hết cho 13
\(A=1+3+3^2+3^3+3^4+...+3^{2015}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{2012}+3^{2013}+3^{2014}+3^{2015}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{2012}\left(1+3+3^2+3^3\right)\)
\(=\left(1+3+3^2+3^3\right)\left(1+3^4+...+3^{2012}\right)\)
\(=40\left(1+3^4+...+3^{2012}\right)\)\(⋮\)\(5\)
\(B=2+2^2+2^3+...+2^{2016}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+..+2^{2013}\left(1+2+2^2+2^3\right)\)
\(=\left(1+2+2^2+2^3\right)\left(2+2^5+...+2^{2013}\right)\)
\(=15\left(2+2^5+...+2^{2013}\right)\)\(⋮\)\(15\)
1. \(A=2^{2016}-1\)
\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)
\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)
16 chia 5 dư 1 nên 16^504 chia 5 dư 1
=> 16^504-1 chia hết cho 5
hay A chia hết cho 5
\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)
lý luận TT trg hợp A chia hết cho 5
(3;5;7)=1 = > A chia hết cho 105
2;3;4 TT ạ !!
A = 3 + 32 + 33 + 34 + ... + 32015 + 32016
A = (3 + 32) + (33 + 34) + ... + (32015 + 32016)
A = 3(1 + 3) + 33(1 + 3) + ... + 32015(1 + 3)
A = 3.4 + 33.4 + ... + 32015.4
A = 4(3 + 33 + ... + 32015)
Vì 4(3 + 33 + ... + 32015) \(⋮\) 4 nên A \(⋮\) 4
Vậy A \(⋮\) 4
A = 3 + 32 + 33 + 34 + ... + 32015 + 32016
A = (3 + 32 + 33) + (34 + 35 + 36) + ... + (32014 + 32015 + 32016)
A = 3(1 + 3 + 32) + 34(1 + 3 + 32) + ... + 32014(1 + 3 + 32)
A = 3.13 + 34.13 + ... + 32014.13
A = 13(3 + 34 + ... + 32014)
Vì 13(3 + 34 + ... + 32014) \(⋮\) 13 nên A \(⋮\) 13
Vậy A \(⋮\) 13