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\(3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=40\left(3+...+3^{2009}\right)⋮40\)
a) \(A=2+2^2+...+2^{120}\)
\(\Rightarrow A=\left(2+2^2\right)+...+\left(2^{119}+2^{120}\right)\)
\(\Rightarrow A=\left(2+2^2\right)+...+2^{118}.\left(2+2^2\right)\)
\(\Rightarrow A=6+...+2^{118}.6\)
\(\Rightarrow A=6.\left(1+...+2^{118}\right)⋮3\Rightarrow A⋮3\left(đpcm\right)\)
b) \(A=2+2^2+...+2^{120}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+2^{117}.\left(2+2^2+2^3\right)\)
\(\Rightarrow A=14+...+2^{117}.14\)
\(\Rightarrow A=14.\left(1+...+2^{117}\right)⋮7\Rightarrow A⋮7\left(đpcm\right)\)
Ta có: A= 2 + 22 + 23 + ... + 260= (2 +22) + (23+ 24) + ... + (259 + 260).
= 2 x (2 + 1) + 23 x (2 + 1) + ... + 259 x (2 + 1).
= 2 x 3 + 23 x 3 + ... + 259 x 3.
= 3 x ( 2 + 23 + ... + 259).
Vì A = 3 x ( 2 + 23 + ... + 259) nên A chia hết cho 3.
A= (2 +22 + 23) + (24 + 25 + 26) + ... + (258 + 259 + 260).
= 2 x (1 + 2 + 22) + 24 x (1 + 2 + 22) + ... + 258 x (1 + 2 + 22).
= 2 x 7 + 24 x 7 + ... + 258 x 7.
= 7 x ( 2 + 24 + ... + 258).
Vì A = 7 x ( 2 + 24 + ... + 258) nên A chia hết cho 7.
A= (2 +22 + 23 + 24) + (25 + 26 + 27 + 28) + ... + (257 + 258 + 259 + 260).
= 2 x (1 + 2 + 22 + 23) + 25 x (1 + 2 + 22 + 23) + ... + 257 x (1 + 2 + 22 + 23).
= 2 x 15 + 25 x 15 + ... + 257 x 15.
= 15 x ( 2 + 24 + ... + 258).
Vì A = 15 x ( 2 + 24 + ... + 258) nên A chia hết cho 15.
1) Ta có : 11a + 22b + 33c
= 11a + 11.2b + 11.3c
= 11.(a + 2b + 3c) \(⋮\)11
=> 11a + 22b + 33c \(⋮\)11
2) 2 + 22 + 23 + ... + 2100
= (2 + 22) + (23 + 24) + ... + (299 + 2100)
= (2 + 22) + 22.(2 + 22) + ... + 298.(2 + 22)
= 6 + 22.6 + ... + 298.6
= 6.(1 + 22 + .. + 298)
= 2.3.(1 + 22 + ... + 298) \(⋮\)3
=> 2 + 22 + 23 + ... + 2100 \(⋮\)3
3) Ta có: abcabc = abc000 + abc
= abc x 1000 + abc
= abc x (1000 + 1)
= abc x 1001
= abc .7. 13.11 (1)
= abc . 7 . 13 . 11 \(⋮\)7
=> abcabc \(⋮\)7
=> Từ (1) ta có : abcabc = abc x 7.11.13 \(⋮\)11
=> abcabc \(⋮\)11
=> Từ (1) ta có : abcabc = abc . 7.11.13 \(⋮\) 13
=> => abcabc \(⋮\)13
1
.\(11a+22b+33c=11\left(a+2b+3c\right)⋮11\)
\(\Rightarrow11a+22b+33c⋮11\left(đpcm\right)\)
hc tốt
A=(2+22+23+24+24)+...+(297+298+299+2100)
A=31+25(2+22+23+24+24)+...+297(2+22+23+24+24)
A=31(1+25+...+297)
=>2+2^2+2^3+.......+2^100) chia hết cho 31
Ta có:
$A=1+2^2+2^4+2^6+...+2^{20}+2^{22}$
$=(1+2^2+2^4)+(2^6+2^8+2^{10})+(2^{12}+2^{14}+2^{16})+(2^{18}+2^{20}+2^{22})$
$=21+2^6\cdot(1+2^2+2^4)+2^{12}\cdot(1+2^2+2^4)+2^{18}\cdot(1+2^2+2^4)$
$=21+2^6\cdot21+2^{12}\cdot21+2^{18}\cdot21$
$=21\cdot(1+2^6+2^{12}+2^{18})$
Vì $21\vdots7$
nên $21\cdot(1+2^6+2^{12}+2^{18})\vdots7$
hay $A\vdots7$ (1)
Lại có:
$A=1+2^2+2^4+2^6+...+2^{20}+2^{22}$
$=(1+2^2+2^4+2^6)+(2^8+2^{10}+2^{12}+2^{14})+(2^{16}+2^{18}+2^{20}+2^{22})$
$=85+2^8\cdot(1+2^2+2^4+2^6)+2^{16}\cdot(1+2^2+2^4+2^6)$
$=85+2^8\cdot85+2^{16}\cdot85$
$=85\cdot(1+2^8+2^{16})$
Vì $85\vdots17$
nên $85\cdot(1+2^8+2^{16})\vdots17$
hay $A\vdots17$ (2)
Mặt khác: $(7,17)=1$ (3)
Từ (1); (2) và (3) $\Rightarrow A\vdots 7\cdot17=119$
$\text{#}Toru$
c)D=4+42+43+44+...+42012
D=(4+42)+(43+44)+...+(42011+42012)
D=4.5+43.5+45.5+...+42011.5
D=5.(4+43+42011)
=>D chia hết cho 5
=>ĐPCM