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Bài 1:
a: \(S=3^0+3^2+3^4+\cdots+3^{2002}\)
=>9S=\(3^2+3^4+3^6+\cdots+3^{2004}\)
=>9S-S=\(3^2+3^4+\cdots+3^{2004}-3^0-3^2-\cdots-3^{2002}\)
=>8S=\(3^{2004}-1\)
=>\(S=\frac{3^{2004}-1}{8}\)
b: \(S=3^0+3^2+3^4+\cdots+3^{2002}\)
\(=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+\cdots+\left(3^{1998}+3^{2000}+3^{2002}\right)\)
\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+\cdots+3^{1998}\left(1+3^2+3^4\right)\)
\(=91\left(1+3^6+\cdots+3^{1998}\right)=7\cdot13\cdot\left(1+3^6+\cdots+3^{1998}\right)\) ⋮7
Bài 2:
b: \(9^{20}=\left(3^2\right)^{20}=3^{40}\)
\(27^{13}=\left(3^3\right)^{13}=3^{39}\)
mà \(3^{40}>3^{39}\) (40>39)
nên \(9^{20}>27^{13}\)
c: \(27^5=\left(3^3\right)^5=3^{15};243^3=\left(3^5\right)^3=3^{15}\)
Do đó: \(27^5=243^3\)
a, Chứng minh rằng A chia hết cho 3
A = 2 + 22 + 23 + .....+ 260
A = ( 2+22 ) + (23 + 24 ) + .....+ (259 + 260 )
A = 2(1+2 ) + 23(1+2) +,...+ 259(1+2)
A = 2.3 + 23.3 + ....+259.3
A = 3(2+23+....+259 ) \(⋮3\)
=> đpcm
chứng minh ằng A chia hết cho 7
A = 2+22 + 23 + .....+ 260
A = ( 2+22 + 23 ) + (24 + 25 + 26) + .... + (258+259+260)
A = 2(1+2 +22 ) +24 (1+2 +22 ) + .... +258(1+2 +22 )
A = 2.7 +24.7 + ....+258.7
A= 7(2+24 ....+258 )\(⋮7\)
=> đpcm
Chứng minh A chia hết cho 15
A = 2 + 22 + 23 + .....+ 260
A = ( 2 + 22 + 23 +24 ) +....+ (257 + 258 + 259 + 260 )
A = 2(1+2+22 + 23 ) + .....+ 257(1+2+22+23)
A = 2.15 + ....+ 257.15
A = 15.(2+...+257) \(⋮15\)
=> đpcm
b,
chứng minh chia hết cho 13
B= 3 + 33 + 35 + + ..........+ 31991
B = (3+33 + 35 ) + (37 + 39 +311 ) + ......+ (31987 + 31989 + 31991 )
B = 3(1+32 +34 ) + 37(1+32 + 34 ) + ....+ 31987(1+32 + 34 )
B = 3.91 + 37.91 + ...+ 31987.91
B = 91(3+37 + ... 31987 )
B = 7.13.(3+37 + ... 31987 ) \(⋮13\)
=> đpcm
chứng minh chia hết cho 41
B = 3+33 + 35 + ...+ 31991
B = (3+33 + 35 + 37 ) + ...(31985 + 31987 + 31989 + 31991 )
B = 3(1+32 + 34 + 36 ) + ...+ 31985(1+32 + 34 + 36)
B = 3. 820 + ...+ 31985.820
B = 820(3+...+31985)
B = 20.41 (3+...+31985) \(⋮41\)
=> đpcm
(-7)90=790=(72)45
3135=(33)45
Vì 72>33=>(-7)90>3135
b/ Ta co : \(3^{15}-9^6=3^{15}-\left(3^2\right)^6=3^{15}-3^{12}=3^{12}.\left(3^3-1\right)=3^{12}.26\)
\(Do\)\(26⋮13\Rightarrow3^{12}.26⋮13\)\(hay\)\(3^{15}-9^6⋮13\)