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\(M=2+2^2+2^3+...+2^{20}\)
\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(M=2\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
\(M=2\cdot15+...+2^{17}\cdot15\)
\(M=15\cdot\left(2+...+2^{17}\right)⋮15\left(đpcm\right)\)
Ta có ;
M = 2 + 22+23+....+220
M = ( 2 + 22+23+24 ) + ....+ ( 217 + 218 + 219 + 220)
M = 2(1 + 2 + 22 + 23)+....+217(1 + 2 + 22 + 23 )
M = 2 . 15 + .... + 217 . 15
Vì 15 chia hết cho 15
Nên 2. 5 + ...+217 . 15
Vậy nên M chia hết cho 15
\(M=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(M=4+13\cdot\left(3^2+3^5+...+3^{98}\right)\)chia 13 dư 4
\(M=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(M=1+40\cdot\left(3+...+3^{97}\right)\)chia 40 dư 1
92.(5x-2)3=37
81.(5x-2)3=2187
(5x-2)3=2187:81
(5x-2)3=27
suy ra :33=27
\(A=2009+2009^2+2009^3+...+2009^{10}\) (có 10 số hạng)
\(A=\left(2009+2009^2\right)+\left(2009^3+2009^4\right)+...+\left(2009^9+2009^{10}\right)\) (có 5 nhóm)
\(A=2009\left(1+2009\right)+2009^3\left(1+2009\right)+...+2009^9\left(1+2009\right)\)
\(A=2009.2010+2009^3.2010+...+2009^9.2010\)
\(A=2010\left(2009+2009^3+...+2009^9\right)\)
Ta thấy: \(2010\left(2009+2009^3+...+2009^9\right)⋮2010\) (Vì \(2010⋮2010\) )
\(\Rightarrow A⋮2010\) (đpcm)
Vậy \(A⋮2010\)
A = (2009 + 20092 + 20093 + 20094 + .... + 200910)
A = [(2009 + 20092) + (20093 + 20094) + ... + (20099 + 200910)]
A = [4038090 + 20092(2009 + 20092) + ... + 20098(2009 + 20092)]
A = [4038090 + 20092.4038090 ... + 20098. 4038090] ⋮ 2010
(4038090 ⋮ 2010)
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a) Ta có: \(B=1+3+3^2+....+3^{2006}\)
\(\Leftrightarrow3B=3+3^2+.....+3^{2006}+3^{2007}\)
\(\Rightarrow3B-B=3^{2007}-1\)
\(\Leftrightarrow B=\dfrac{3^{2007}-1}{2}\)
Vậy \(B=\dfrac{2^{2007}-1}{2}\)
b) Ta có: \(A=3^{2007}-1=\left(3-1\right)\left(3^{2006}+3^{2005}+.......+3+1\right)\)
\(\Leftrightarrow A=2\left(3^{2006}+3^{2005}+....+3+1\right)\) luôn chia hết cho 2
Vậy \(A=\left(3^{2007}-1\right)⋮2\)
a) \(B=1+3+3^2+3^3+3^4+.......+3^{2006}\)
\(\Leftrightarrow3B=3+3^2+3^3+3^4+.......+3^{2007}\)
\(\Leftrightarrow3B-B=\left(3+3^2+3^3+3^4+.......+3^{2007}\right)-\left(1+3+3^2+3^3+3^4+.......+3^{2006}\right)\)
\(\Leftrightarrow2B=3^{2007}-1\)
\(\Leftrightarrow B=\dfrac{3^{2007}-1}{2}\)
Vậy \(B=\dfrac{3^{2007}-1}{2}\)
ý a)là mình biết làm rồi có phải như vậy không
C = 3 + 32 + 33 + .......3100
=(3+32+33+34)+(35+36+37+38)+......+(397+398+399+3100)
=3.(1+3+32+33)+35(1+3+32+33)+.....+397.(1+3+32+33)
=3.40 + 35.40 +.......+397.40
=40.(3 + 35+ ...+397)
Suy ra C chia hết cho 40
\(C=3+3^2+3^3+....+3^{100}\)
\(C=\left(3+3^2+3^3+3^4\right)+....+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(C=120+....+3^{97}.\left(3+3^2+3^3+3^4\right)\)
\(C=120+....+3^{97}.120\)
\(\Rightarrow C⋮40\)
MÌNH CHỈ GIẢI ĐƯỢC MỘT BÀI THÔI NHÉ !