Chung minh tong sau chia het cho 7
A=2^1+2^2+2^3+...+2^59+2^60
Chung minh tong sau chia het cho7
A=2^1+2^2+2^3+....+2^59+2^60
A = (21 + 22 + 23 + ... + 259 + 260)
A = 20(21 + 22 + 23) + 23(21 + 22 + 23) + ... + 257(21 + 22 + 23)
A = (21 + 22 + 23) + (20 + 23 + ... + 257)
A = 14(20 + 23 + ... + 257) chia hết cho 7
=> A chia hết cho 7.
A=2(1+2+4)+24(1+2+4)+27(1+2+4)+...+258(1+2+4)
A=(1+2+4)(2+24+27+...+258)
A=7.(2+24+27+...+258) chia hết cho 7
A= (2^1 + 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)
= 2.7 + 2^4.7 +...+ 2^58.7
= 7(2 + 2^5 +...+ 2^58)
Vậy A chia hết cho 7
Chung minh tong s=1+2+2^2+2^3+.....+2^59 chia het cho 3
Ta có S=1+2+22+23+...+259
\(\Rightarrow\)2S=2+22+23+24+...+260
\(\Rightarrow\)2S-S=260-1
do 2 chia 3 dư 1 \(\Rightarrow\)260 chia 3 dư 160\(\Rightarrow\)260 chia 3 dư 1
\(\Rightarrow\)260 -1 \(⋮\)3
Hay S\(⋮\)3 (dpcm)
\(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=3+2^2\left(1+2\right)+...+2^{58}\left(1+2\right)\)
\(=3+2^2\times3+...+2^{58}\times3\)
\(=3\times\left(1+2^2+...+2^{58}\right)⋮3\)
Vậy \(S⋮3\)
Cho A= 2 + 22 + 23 + .... + 259 + 260
Chung minh A chia het cho 15
\(A=2+2^2+2^3+...+2^{59}+2^{60}\)
\(=2.\left(1+2+2^2+2^3\right)+2^5.\left(1+2+2^2+2^3\right)+...+2^{57}.\left(1+2+2^2+2^3\right)\)
\(=2.15+2^5.15+....+5^{57}.15\)
\(=15.\left(2.2^5...2^{57}\right)⋮15\)
Vậy:\(A⋮15\)
cho A=2+2^2+2^3+....+2^59+2^60
chung minh Acgia het cho 21
chung minh A= 2 + 2^2 +2^3 +2^4 +.........+2^60 chia het cho 7
tim so tu nhien n de : n+4 chia het cho n+1
chung minh ( 1+2 +2^2 +2^3+2^4+2^5+2^6+2^7) chia het cho 3
1. A = 2 + 22 + 23 + 24 + ... + 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 = ( 2 + 24 + ... + 258 ) . 7 => A \(⋮\)7
Vậy ...
2.Ta có : \(n+4⋮n+1\)
Mà : \(n+1⋮n+1\)
\(\Rightarrow\left(n+4\right)-\left(n+1\right)⋮n+1\Rightarrow n+4-n-1⋮n+1\)
\(\Rightarrow3⋮n+1\Rightarrow n+1\in\left\{1;3\right\}\)
\(\Rightarrow n\in\left\{0;2\right\}\)
3. Đặt B = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
B = ( 1 + 2 ) + ( 22 + 23 ) + ( 24 + 25 ) + ( 26 + 27 )
B = ( 1 + 2 ) + 22 ( 1 + 2 ) + 24 ( 1 + 2 ) + 26 ( 1 + 2 )
B = 1 . 3 + 22 . 3 + 24 . 3 + 26 . 3
B = ( 1 + 22 + 24 + 26 ) . 3 \(\Rightarrow\) B \(⋮\)3
Vậy ...
C=1+2+22+............259
a)chung minh c va 260 la hai so tu nhien lien tiep
b)chung min c chia het cho 3
ta có \(2C=2+2^2+2^3+...+2^{60}\)
=> \(2C-C=2+2^2+2^3+...+2^{60}-1-2-2^2-...-2^{59}=2^{60}-1\)
=> \(C=2^{60}-1\)
=> C và \(2^{60}\) là 2 số tự nhiên liên tiếp (ĐPCM)
cau lam on lam luon cau b gium minh nha
chứng minh ý a 8mu10 cong 2 mũ 20 chia hết cho41
chứng minh ý b 31 mu36 nhận 36 -313 mu 5 nhận 299 chia hết ch
chung minhý c 3 mu n công 3 cộng 2 mu n công 3 công 2 mu n công 2 công 2 mu n công 2 chia hết 6
chung minh y d d=2 cong 2 mu 2 cong 2mu 3 cong 2 mu 4 cong 2 mu 5 cong 2 mu 6 cong ....cong 2 mu 58 cong 2 mu 59 cong 2 mu 60 chia het cho 31
chung minh y e, e= 1cong 3 cong 3 mu 2 cong 3 mu 3 cong..cong 3 mu 98 cong 3 mu 99 chung minh e chia het cho5 chia het cho11
CHO A= 3+3MU2+3mu3+3mu4+...+3mu2017 a) tim so tu nhien N biet 2A +3 = 3n b)tim chu so tan cung cua A
chung minh C= 21+22+...+260 chia het cho 3; chia het cho 7
cho A = 2 + 2^2 + 2^3 + ... + 2^60
chung to tong A chia het cho 2, cho 3, cho 7, cho 14