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\(C=2+2^2+2^3+......+2^{100}⋮31\)
\(C=2.\left(1+2+2^2+2^3+2^4\right)+2^{95}.\left(1+2+2^2+2^3+2^4\right)\)
\(C=2.31+.......+2^{95}+31\)
\(C=31.\left(2+2^{95}\right)⋮31\)
\(\Rightarrow C⋮31\)
a, Ta co : M= ( 1 +4 + 42 ) + ( 43 + 44 + 45 ) +.......................+ ( 42010 + 42011 +42012 )
M = 1. (1+4+16 ) +43. (1+4+16 ) +.........................+ 42010. ( 1+4 +16
M = 1, 21 + 43. 21 +..............................................+ 42010 .21
M= 21.(1+43+.................................... + 42010 ) CHIA HẾT 21
TƯƠNG TƯ
a) S = 5 + 52 + 53 + ... + 5100
=> S = ( 5 + 52 ) + ( 53 + 54 ) + ... + ( 599 + 5100 )
=> S = 5( 1 + 5 ) + 53( 1 + 5 ) + ... + 599( 1 + 5 )
=> S = 5 . 6 + 53 . 6 + ... + 599 . 6
=> S = ( 5 + 53 + ... + 599 ) . 6 chia hết cho 6
=> S chia hết cho 6
b) S1 = 2 + 22 + 23 + ... + 2100
=> S1 = ( 2 + 22 + 23 + 24 + 25 ) + ... + ( 296 + 297 + 298 + 299 + 2100 )
=> S1 = 2( 1 + 2 + 22 + 23 + 24 ) + ... +296( 1 + 2 + 22 + 23 + 24 )
=> S1 = 2 . 31 + ... + 296 . 31
=> S1 = ( 2 + ... + 296 ) . 31 chia hết cho 31
=> S1 chia hết cho 31
c) S2 = 165 + 215
=> S2 = ( 24 )5 + 215
=> S2 = 220 + 215
=> S2 = 220( 1 + 25 )
=> S2 = 220 . 33 chia hết cho 33
=> S2 chia hết cho 33
\(S1=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^{99}.\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{99}.6\)
\(=6.\left(5+5^3+...+5^{99}\right)⋮6\)
câu b tương tự
\(S3=16^5+21^5\)
vì 16+21=33 chia hết cho 33
=>165+215 chia hết cho 33
P/S: theo công thức:(n+m chia hết cho a=> nb+mb chia hết cho a)
S1 = 5+52+53+...+599+5100
=5. (1+5)+53 . (1+5) + ... + 599.(1+5)
= 5.6 +53.6+..+ 599.6
=6.(5+53 + ... +599):6
vậy x = ...
b)2+22+23+...+299+2100
=2.(1+2)+23.(1+2) + ... + 299.(1+2)
=2.3+23+..+299):3
= ....
c)165+215
vì 16+21 chia hế 33 nên
theo công thức(n+m chia hết cho a=(nb+mb)
\(a)\) Đặt \(A=5+5^2+5^3+5^4+...+5^{99}+5^{100}\)ta có :
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(A=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(A=5.6+5^3.6+...+5^{99}.6\)
\(A=6.\left(5+5^3+...+5^{99}\right)\) \(⋮\) \(6\)
Vậy \(A⋮6\)
\(b)\) Đặt \(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}\) ta có :
\(B=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(B=2\left(1+2+4+8+16\right)+...+2^{96}\left(1+2+4+8+16\right)\)
\(B=2.31+...+2^{96}.31\)
\(B=31.\left(2+2^6+...+2^{96}\right)\) \(⋮\) \(31\)
Vậy \(B⋮31\)
Năm mới zui zẻ ^^
ta có: 2+2^2+2^3+...+2^100
(=) (2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^97+2^98+2^99+2^100)
(=) 30 +2^4.(2+2^2+2^3+2^4)+...+2^96.(2+2^2+2^3+2^4)
(=) 30+2^4.30+...+2^96.30
(=) 30.(1+2^4+...+2^96)
Vì 30\(⋮\)5=> 2+2^2+2^3+2^4+...+2^100\(⋮\)5
ta có 2+2^2+2^3+....+2^100
(=) (2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)+....+(2^95+2^96+2^97+2^98+2^100)
(=) 62+2^5.(2+2^2+2^3+2^4+2^5)+...+2^94.(2+2^2+2^3+2^4+2^5)
(=) 62+2^5.62+...+2^94.62
Vì 62\(⋮\)31 => 2+2^2+2^3+...+2^100\(⋮\)31
ai nhanh mình k 3 cái