Chứng minh 2^28−1⋮29
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\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{27}+2^{28}+2^{29}\right)\\ A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{27}\left(1+2+2^2\right)\\ A=\left(1+2+2^2\right)\left(1+2^3+...+2^{27}\right)\\ A=7\left(1+2^3+...+2^{27}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 + 2 + 23 + 24 +...+ 228 + 229
= 20 + 21 + 23 + 24 +...+ 228 + 229
= (20 + 21) + (23 + 24) +...+ (228 + 229)
= 20(20 + 21) + 23(20 + 21) +...+ 228(20 + 21)
= 20 . 3 + 23 . 3 +...+ 228 . 3
= (20 + 23 + 26 +...+ 228) . 3 chia hết cho 3
1 + 2 + 22 + 23 +...+ 228 + 229
= 20 + 21 + 22 + 23 +...+ 228 + 229
= (20 + 21) + (22 + 23) +...+ (228 + 229)
= 20(20 + 21) + 22(20 + 21) +...+ 228(20 + 21)
= 20 . 3 + 22 . 3 +...+ 228 . 3
= (20 + 22 + 24 +...+ 228) . 3 chia hết cho 3
Hi hi. Mình nhầm tí.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=3^1+3^2+3^3+...+3^{28}+3^{29}+3^{30}\)
\(M=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)...+3^{28}\left(1+3+3^2\right)\)
\(M=3.13+3^4.13...+3^{28}.13\)
\(M=13.\left(3+3^4...+3^{28}\right)⋮13\)
\(\Rightarrow dpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(2^{28}-1=\left(2^{14}\right)^2-1\equiv1^2-1\left(mod9\right)\)
Vậy \(2^{28}-1⋮29\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2^1+2^2+2^3+2^4+2^5+2^6+..+2^{28}+2^{29}+2^{30}\)
\(S=2.\left(1+2+2^2\right)+2^4.\left(1+2+2^2\right)+...+2^{28}.\left(1+2+2^2\right)\)
\(S=\left(1+2+2^2\right).\left(2+2^4+...+2^{28}\right)\)
\(S=7.\left(2+2^4+...+2^{28}\right)\)
⇒ \(S⋮7\) ( điều phải chứng minh )
![](https://rs.olm.vn/images/avt/0.png?1311)
G = 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210
2.G = 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211
2G - G = (22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211) - (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210)
G = 22 + 23 + 24 +25 + 26 + 27 + 28 + 29 + 210 + 211 - 21 -22 -23 -24 - 25 - 26 - 27 - 28 - 29 - 210
G = (22 -22) +(23 - 23) + (24 - 24) + (25 -25) + (26 - 26) +(27 - 27) +(28 -28) + (29 - 29) + (210 - 210) + (211 - 21)
G = 211 - 2
G = 2048 - 2 (đpcm)
b,
G = 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210
D = 2.(1+ 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29)
Vì 2 ⋮ 2 nên D = 2.(1+2+22+23+24+25+26+27+28+29)⋮2 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^6-25^3=\left(5^2\right)^3-25^3=25^3-25^3=0\)
\(\Rightarrow\frac{\left(1^6-29^3\right)\left(2^6-28^3\right)\left(3^6-27^3\right)\left(4^6-26^3\right)\left(5^6-25^3\right).....\left(10^6-20^3\right)}{\left(1^6+29^3\right)\left(2^6+28^3\right)\left(3^6+27^3\right)\left(4^6+26^3\right)\left(5^6+25^3\right).....\left(10^6+20^3\right)}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(M=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)\)
\(=3\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{28}.\left(1+3+3^2\right)\)
\(=13\left(3+3^4+...+3^{28}\right)⋮13\Rightarrow M⋮13\)
M = 31 + 32 + 33 +...+ 328 + 329 + 330
M = ( 31 + 32 + 33) + ...+ ( 328 + 329 + 330 )
M = 3(1 + 3 + 32 ) +...+ 328( 1 + 3 + 32)
M = 3 .13 +...+ 328.13
\(\Rightarrow M⋮13\)(đpcm)
!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
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