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S = 1 + 3 + 32 + 33 + ... + 38 + 39
S = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 38 + 39 )
S = 4 + ( 1 . 32 + 3 .32 ) + .. + ( 1. 38 + 3 . 38 )
S = 4 + 4 .32 + .. + 4 . 38
S = 4 ( 1 + 32 + ... + 38 ) \(⋮\)4
Vậy S \(⋮\)4 ( đpcm )
Học tốt
#Dương
S = 1 + 3 + 32 + 33 + 34+35+ 36 + 37 + 38+39
S=( 1 + 3)+(32 + 33)+(34+35)+(36 + 37)+(38+39)
s=4+32.(3+1)+32.(3+1)+34.(3+1)+36.(3+1)+38.(3+1)
S=4.(1+32+34+36+38)
CHIA HẾT CHO 4
a, S=1+2^7+(2+2^2)+(2^3+2^4)+(2^5+2^6)
S=1+128+2*3+(2^3*1+2^3*2)+(2^5*1+2^5*2)
S=129+2*3+2^3*(1+2)+2^5*(1+2)
S=3*43+2*3+2^3*3+2^5*3
S=3*(43+2+2^3+2^5)chia hết cho 3 nên S chia hết cho 3
c) S = ( -2 ) + 4+ ( -6 ) + 8 + ... + ( -2002 ) + 2004
S = [ (-2)+4] + [ (-6) + 8 ] + ... + [ (-2002) + 2004 ]
S = 2 + 2 + 2 + ... + 2 ( 501 số hạng 2 )
S = 2*501
S = 1002
a: \(S=\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(=4\left(1+3^2+3^4+...+3^8\right)⋮4\)
b: \(S=\left(1+2\right)+2^2\left(1+2\right)+...+2^8\left(1+2\right)\)
\(=3\left(1+2^2+...+2^8\right)⋮3\)
Ta có ;
S = 1 + 2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7
= ( 1 + 2 ) + ( 2 2 + 2 3 ) + ( 2 4 + 2 5 ) + ( 2 6 + 2 7 )
= ( 1 + 2 ) + 2 2 ( 1 + 2 ) + 2 4 ( 1 + 2 ) + 2 6 ( 1 + 2 )
= 3 + 2 2 .3 + 2 4 .3 + 2 6 .3
= 3 . ( 1 + 2 2 + 2 4 + 2 6 ) chia hết cho 3 ( Vì 3 chia hết cho 3 )
A = 3 + 3 2 + 3 3 + ..... + 3 9 + 3 10
= ( 3 + 3 2 ) + ( 3 3 + 3 4 ) .... + ( 3 9 + 3 10 )
= 3 ( 1 + 3 ) + 3 3 . ( 1 + 3 ) + .... + 3 9 ( 1 + 3 )
= 3 . 4 + 3 3 . 4 + .... + 3 9 . 4
= 4 . ( 3 + 33 + ... + 3 9 ) chia hết cho 4 ( Do 4 chia hết cho 4 )
\(S=\left(1+2\right)+\left(2^2+2^3\right)+\left(2^4+2^5\right)+\left(2^6+2^7\right)\)
\(S=3+3\cdot2^2+3\cdot2^4+3\cdot2^6=3\left(1+2^2+2^4+2^6\right)⋮3\)
\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^9+3^{10}\right)\)
\(A=4\cdot3+4\cdot3^3+...+4\cdot3^9=4\cdot\left(3+3^3+...+3^9\right)⋮4\)
a/ Ta có :
\(S=1+3+3^2+........+3^{2017}\)
\(\Leftrightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+......+\left(3^{2016}+3^{2017}\right)\)
\(\Leftrightarrow S=1\left(1+3\right)+3^2\left(1+3\right)+......+3^{2016}\left(1+3\right)\)
\(\Leftrightarrow S=1.4+3^2.4+........+3^{2016}.4\)
\(\Leftrightarrow S=4\left(1+3^2+......+3^{2016}\right)⋮4\left(đpcm\right)\)
b/ \(S=1+3+..........+3^{2017}\)
\(\Leftrightarrow3S=3+3^2+.........+3^{2017}+3^{2018}\)
\(\Leftrightarrow3S-S=\left(3+3^2+..........+3^{2018}\right)-\left(1+3+.....+3^{2017}\right)\)
\(\Leftrightarrow2S=3^{2018}-1\)
\(\Leftrightarrow S=\dfrac{3^{2018}-1}{2}\)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4