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\(A=7+7^2+7^3+...+7^{120}\)
\(A=\left(7+7^2+7^3\right)+...+\left(7^{118}+7^{119}+7^{120}\right)\)
\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)
\(A=7.57+7^4.57+...+7^{118}.57\)
\(A=57\left(7+7^4+...+7^{118}\right)\)
\(\Rightarrow A⋮57\)
A = (2 + 2^2) + (2^3+ 2^4) +...... + (2^119 + 2^120)
A= (2.1+2.2) + (2^3.1 + 2^3.2) + ...... + (2^119.1 + 2^119.2)
A = 2.3 + 2^3.3 + ...... + 2^119.3
A = 3.(2+2^3+......+2^119)
Chia hết cho 3
A = (2 + 2^2 + 2^3) +...... + (2^118 + 2^119 + 2^120)
A = (2.1 + 2.2 + 2.4) + ....... + (2^118.1 + 2^118.2 + 2^118.4)
A = 2.(1+2+4) + ...... + 2^118.(1 + 2 + 4)
A= 7.(2 + 2^4 + ...... + 2^118)
Chia hết cho 7
+)A=2^1+2^2+2^3+2^4+...+2^2010
=>A=(2^1+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^2009+2^2010)
=>A=6+2^2.(2+2^2)+2^4.(2+2^2)+...+2^2008(2+2^2)
=>A=6+2^2.6+2^4.6+...+2^2008.6
=>A=6.(1+2^2+2^4+...+2^2008)
=>A=3.2.(1+2^2+2^4+...+2^2008)
=>A chia hết cho 3
A=2+2^2+2^3+2^4+...+2^2010
A=(2+2^2+2^3)+(2^4+2^5+2^6)+(2^7+2^8+2^9)+...+(2^2008+2^2009+2^2010)
A=2.(1+1+2^2)+2^4(1+2+2^2)+2^7.(1+2+2^4)+...+2^2008.(1+2+2^2)
A=2.7+2^4.7+2^7.7+...+2^2008.7
A=7.(2+2^4+2^7+...+2^2008)
=> A chia hết cho 7
các phần khác làm tương tự
A = 21 + 22 + 23 + 24 + .... + 22009 + 22010
=> A = ( 21 + 22 ) + ( 23 + 24 ) + .... + ( 22009 + 22010 )
=> A = 21.( 1 + 2 ) + 23.( 1 + 2 ) + .... + 22009.( 1 + 2 )
=> A = 21.3 + 23.3 + .... + 22009.3
=> A = 3.( 21 + 23 + .... + 22009 )
Vì 3 ⋮ 3 => A ⋮ 3 ( đpcm )
A = 21 + 22 + 23 + 24 + 25 + 26 + .... + 22007 + 22008 + 22009
=> A = ( 21 + 22 + 23 ) + ( 24 + 25 + 26 ) + .... + ( 22007 + 22008 + 22009 )
=> A = 21.( 1 + 2 + 2.2 ) + 24.( 1 + 2 + 2.2 ) + .... + 22007.( 1 + 2 + 2.2 )
=> A = 21.7 + 24.7 + .... + 22007.7
=> A = 7.( 21 + 24 + .... + 22007 )
Vì 7 ⋮ 7 => A ⋮ 7 ( đpcm )
Các ý sau tương tự .
\(A=\left(2+2^2+2^3+2^4\right)+....+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(A=30+...+2^{16}.\left(2+2^2+2^3+2^4\right)\)
\(A=30+...+2^{16}.30\)
\(A=30.\left(1+...+2^{16}\right)⋮5\)
B tương tự ( 57=3.19)
cm tổng đó chia hết cho 3 và 19 là đc =)
\(A=7+7^2+7^3+...+7^{2016}\)
\(A=\left(7+7^2+7^3\right)+\left(7^4+7^5+7^6\right)+...+\left(7^{2014}+7^{2015}+7^{2016}\right)\)
\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{2014}\left(1+7+7^2\right)\)
\(A=7.57+7^4.57+...+7^{2014}.57\)
\(A=\left(7+7^4+...+7^{2014}\right).57⋮57\) ( đpcm )
Ta có :
\(A=7\left(1+7+7^2\right)+.....+7^{2014}\left(1+7+7^2\right)\)
\(\Rightarrow A=7.57+....+7^{2014}.57\)
\(\Rightarrow A=57.\left(7+....+7^{2014}\right)\)
=> A chia hêt cho 57
Gọi phần a, là A,ta có:
A=1+4+42+43+...+42000
4.A=4.(1+4+42+...+42000)
4.A=4+42+43+44+...+42001
4.A-A=(4+42+43+...+42001)-(1+4+42+...+42000)
3.A=4+42+43+...+42001 -1-4-42-...-42000
3.A=42001-1
A=(42001-1):3
K CHO MIK NHÉ !
A = 7 + 72 + 73 + .... + 72016 có (2016 - 1) : 1 + 1 = 2016 số hạng
A = (7 + 72 + 73) + ... + (72014 + 72015 + 72016)
A = 7 . (1 + 7 + 72) + .... + 72014 . (1 + 7 + 72)
A = 7 . (1 + 7 + 49) + .... + 72014 . (1 + 7+ 49)
A = 7 . 57 + ... + 72014 . 57
A = 57 . (7 + ... + 72014) chia hết cho 57
=> A chia hết cho 57 (ĐPCM)
Ủng hộ mk nha !!! ^_^
A = 7 + 72 + 73 +.....+ 72016
A = (7 + 72 + 73) + (74 + 75 + 76) +....+ (72014 + 72015 + 72016)
A = 7(1+7+72) + 74(1+7+72) +....+ 72014(1+7+72)
A = 7.57 + 74.57 +.....+ 72014.57
A = (7 + 74 +....+ 72014).57 chia hết cho 57 (Đpcm)
a
M=(7+7^2)+(7^3+7^4)+...+(7^59+7^60)
=7.(7+1)+7^3.(7+1)+...+7^59+(7+1)
=7.8+7^3.8+...+7^59+8
=>M chia hết cho8
\(a,A=7^{15}+7^{16}+7^{17}\)
\(A=7^{15}\left(1+7+7^2\right)\)
\(A=7^{15}.57\)
Ta có :
\(A=7^{15}.57⋮57\)
\(\Rightarrow A⋮57\)
\(b,B=2+2^2+2^3+....+2^{60}\)
\(B=\left(2+2^2+2^3\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(B=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(B=2.7+...+2^{58}.7\)
\(B=7\left(2+2^4+....+2^{58}\right)\)
Ta có :
\(B=7\left(2+2^4+....+2^{58}\right)⋮7\)
\(\Rightarrow B⋮7\)
\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)=7.57+7^4.57+...+7^{118}.57=57\left(7+7^4+...+7^{118}\right)⋮57\)
Lời giải:
$A=(7+7^2+7^3)+(7^4+7^5+7^6)+....+(7^{118}+7^{119}+7^{120})$
$=7(1+7+7^2)+7^4(1+7+7^2)+...+7^{118}(1+7+7^2)$
$=7.57+7^4.57+...+7^{118}.57$
$=57(7+7^4+...+7^{118})\vdots 57$
Ta có đpcm.