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a) tổng S bằng
(2014+4).671:2=677 039
b)n.(n+2013) để mọi số tự nhiên n mà tổng trên chia hét cho 2 thì n=2n
→2n.(n+2013)\(⋮̸\)2
C)M=2+22+23+...+220
=(2+22+23+24)+...+(217+218+219+220)
=(2+22+23+24)+...+(216.2+216.22+216+23+216.24)
=30.1+...+216.(2+22+23+24)
=30.1+...+216.30
=30(1+25+29+213+216)\(⋮\)5
c, M= 2 + 22 + 23 +........220
Nhận xét: 2+ 22 + 23 + 24 = 30; 30 chia hết cho 5
Khi đó: M = ( 2+22 + 23 + 24 ) + (25 + 26 + 27 + 28)+.....+ (217+218+219+220)
= ( 2+22 + 23 + 24 ) + 24. ( 2+22 + 23 + 24 ) +...........+216 .( 2+22 + 23 + 24 )
= 30+24 .30 + 28. 30 +.........+ 216.30
= 30.(24 + 28 +.........+216) chia hết cho 5 và 30 chia hết cho 5
Vậy M chia hết cho 5
\(M=2+2^3+2^5+2^7+....+2^{51}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)
\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)
\(=10+2^4.10+...+2^{48}.10\)
\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)
\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)
\(M=2+2^3+2^5+2^7+....+2^{51}.\)
\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)
\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)
\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)
\(=12+2^4.42+....+2^{46}.42\)
\(=12+7.3.2\left(2^4+...+2^{46}\right)\)
\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)
\(=10+7.3.2\left(2^4+....+2^{46}\right)\)
Ta có: \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7
Suy M không chia hết cho 7
Ta có: \(2^{n+3}+5^n-2^{n+1}+5^{n+1}=\left(2^{n+3}+2^{n+1}\right)+\left(5^n+5^{n+1}\right)\)
\(=2^n\left(2^3-2\right)+5^n\left(1+5\right)=2^n.6+5^n.6=6.\left(2^n+5^n\right)⋮6\left(đpcm\right)\)
Kb với mình nhé ~_~
2^n+3+5^n-2^n+1+5^n+1
= (2^n+3-2^n+1) + (5^n+5^n+1)
= 2^n.(2^3-2)+5^n.(5+1)
= 2^n.6+5^n.6 = 6.(2^n+5^n) chia hết cho 6
k mk nha
a, \(10^m-1⋮19,19⋮19\)
\(\Rightarrow\left(10^m-1\right)\left(10^m+1\right)+19⋮19\)
\(\Rightarrow10^{2m}-1+19⋮19\Rightarrow10^{2m}+18⋮19\)
\(b,\)Ta có : \(3+3^2+3^3+3^4+...+3^{23}+3^{24}+3^{25}\)
\(=3+\left(3^2+3^3+3^4\right)+...+\left(3^{23}+3^{24}+3^{25}\right)\)
\(=3+3\left(3+3^2+3^3\right)+...+3^{22}\left(3+3^2+3^3\right)\)
\(=3+3.39+...+3^{22}.39\)
\(=3+39\left(3+...+3^{22}\right)\)
Suy ra : B chia 39 dư 3
Vậy : B không chia hết cho 39
S = 1 + 2 + 22 + 23 + ... + 220 + 221 (có 22 số; 22 chia hết cho 2)
S = (1 + 2) + (22 + 23) + ... + (220 + 221)
S = 3 + 22.(1 + 2) + ... + 220.(1 + 2)
S = 3 + 22.3 + ... + 220.3
S = 3.(1 + 22 + ... + 220) chia hết cho 3 (đpcm)
\(S=1+2+2^2+2^3+....+2^{21}\)
\(=\left(1+2\right)+2^2\left(1+2\right)+2^4\left(1+2\right)+......+2^{20}\left(1+2\right)\)
\(=\left(1+2\right)\left(1+2^2+2^4+.....+2^{20}\right)\)
\(=3\left(1+2^2+2^4+....+2^{20}\right)\)
Chia hết cho 3
Ta có :
\(M=2+2^2+2^3+...+2^{20}\)
\(\Rightarrow M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(\Rightarrow M=2\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
\(\Rightarrow M=2.15+...+2^{17}.15\)
\(\Rightarrow M=15\left(2+...+2^{17}\right)\)
\(\Rightarrow M⋮15\)
\(\RightarrowĐPCM\)
viết M dưới dạng:
M=\(2.(1+2+2^2+2^3)+2^5.(1+2+2^2+2^3)+...\)
M=\(2.15+2^5.15+...\)
\(=>\) M chia hết cho 15
Ta có:
88 + 220
= (23)8 + 220
= 23 . 8 + 220
= 220 . 24 + 220
= 220 . (16 + 1)
= 220 . 17 \(⋮\) 17(đpcm)
A=4+(22+23+24+...+220)
A-4=22+23+24+...+220
2(A-4)=23+24+25+...+221
A-4=2(A-4)-(A-4)=(23+24+25+...+221)-(22+23+24+...+220)
A-4=(23-23)+(24-24)+(25-25)+...+(220-220)+(221-22)
A-4=221-4
A =221-4+4
A =221
Bạn làm tiếp nha .
M=2+22+...+220
2M=22+23+...+221
=>2M-M=(22+23+...+221)-(2+22+...+220 )
=>M=221-2=2097150 chia hết cho 5
M = 2+22+23+24+....+220
M=(2+22+23+24)+24x(2+22+23+24)+....+216x(2+22+23+24)
M=30+24x30+....+216x30
M=30x(1+24+.....+216)
mà 30 chia hết cho 5
=>30x(1+24+......+216) chia hết cho 5
=>M chia hết cho 5
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k mình nha