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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/ \(S=1+3+..........+3^{2017}\)
\(\Leftrightarrow3S=3+3^2+.........+3^{2017}+3^{2018}\)
\(\Leftrightarrow3S-S=\left(3+3^2+..........+3^{2018}\right)-\left(1+3+.....+3^{2017}\right)\)
\(\Leftrightarrow2S=3^{2018}-1\)
\(\Leftrightarrow S=\dfrac{3^{2018}-1}{2}\)
1, S = 2+22 + 23 + ....+ 260
a, chứng tỏ S chia hết cho 3
S = 2+22 + 23 + ....+ 260
S = (2+22 ) + (23 + 24 ) + ....+ (259 + 260)
S = 2(1+2 ) + 23(1+2 ) + ....+ 259(1+2)
S = 2.3 + 23 .3 + ....+ 259 .3
S = 3(2+23 + ...+259 ) \(⋮\) 3
=> đpcm
b, chứng tỏ S chia hết cho 7
S = 2+22 + 23 + ....+ 260
S = (2+22 + 23 ) + ....+ ( 258 + 259 + 260)
S = 2(1+2+22 ) + ....+ 258(1+2+22 )
S = 2.7 + ....+ 258 .7
S= 7(2+...+258)\(⋮\) 7
=> đpcm
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: \(S=\left(1+2\right)+2^2\left(1+2\right)+...+2^8\left(1+2\right)\)
\(=3\left(1+2^2+...+2^8\right)⋮3\)