Chứng minh S=3+3^2+3^3+...3^2007chia hết cho 13
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a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{19}\right)⋮7\)
B = (1 + 3) + (32+33)+.....+(389+390)
= 4 + 32 .(1 + 3) + .....+390.(1+3)
= 1 .4 + 32.4 + ..... +390.4
= 4.(1 + 32 + .... +390) chia hết cho 4
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
SCSH: (32015- 1) : 2 = 0
Tổng: (32015+ 1) : 2 = 2
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
S=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^999+3^1000+3^1001)
S=1x(1+3+9)+3^3x(1+3+9)+...+3^999x(1+3+9)
S=1x13+3^3x13+...+3^999x13
S=13x(1+3^3+...+3^999)
Vậy S chia hết cho 13
S=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^999+3^1000+3^1001)
S=1x(1+3+9)+3^3x(1+3+9)+...+3^999x(1+3+9)
S=1x13+3^3x13+...+3^999x13
S=13x(1+3^3+...+3^999)
Vậy S chia hết cho 13
Có: 3(1+3)+3^3(1+3)+.....+3^59(1+3)
=3.4+3^3.4+.....+3^59.4
=>S : hết cho 4
Có: 3(1+3+9)+3^4(1+3+9)+.....+3^58(1+3+9)
=3.13+3^4.13+.....+3^58.13
=>S : hết cho 13
tick cho mình đi !
Ta có : S = 3 + 32 + 33 + ....+ 32007
\(\Rightarrow S=\left(3+3^2+3^3\right)+....+\left(3^{2005}+2^{2006}+2^{2007}\right)\)
\(\Rightarrow S=3\left(1+3+3^2\right)+.....+3^{2005}\left(1+3+3^2\right)\)
\(\Rightarrow S=3\cdot13+....+3^{2005}\cdot13\)
\(\Rightarrow S=13\cdot\left(3+....+2005\right)\)
\(\Rightarrow S\) chia hết cho 13
đúng nha !!!