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A = 3 + 32 + 33 + 34 +..... + 32015 + 32016
= (3 + 32 + 33) + (34+ 35 + 36 ) +.....+ (32014 + 32015 + 32016)
= 3(1 + 3 + 32) + 34(1 + 3 + 32) + .....+ 32014(1 + 3 + 32)
= 13(3 + 34 + ....+ 32014) \(⋮13\)
A = 3 + 32 + 33 + 34 +..... + 32015 + 32016
= (3 + 32) + (33 + 34) + .... + (32015 + 32016)
= 3(1 + 3) + 33(1 + 3) + .... + 32015(1 + 3)
= 4(3 + 33 + .... + 32015) \(⋮4\)
\(A=3+3^3+3^5+3^7+...+3^{2015}⋮13and41\)
\(A=\left(3+3^2+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{2011}+3^{2013}+3^{2015}\right)\)
\(A=3.\left(1+3^2+3^4\right)+3^7.\left(1+3^2+3^4\right)+...+3^{2011}.\left(1+3^2+3^4\right)\)
\(A=3.91+3^7.91+...+3^{2011}.91\)
\(A=3.7.13+3^7.7.13+...+3^{2011}.7.13\)
\(A=13.\left(3.7+3^7.7+...+3^{2011}.7\right)\)
\(forA=13.\left(3.7+3^7.7+...+3^{2011}.7\right)soA⋮13\)
\(A=\left(3+3^3+3^5+3^7\right)+...+\left(3^{2009}+3^{2011}+3^{2013}+3^{2015}\right)\)
\(A=3.\left(1+3^2+3^4+3^6\right)+...+3^{2009}\left(1+3^2+3^4+3^6\right)\)
\(A=3.820+...+3^{2009}.820\)
\(A=3.20.41+...+3^{2009}3.20.41\)
\(A=41.\left(3.20+...+3^{2009}.20\right)\)
\(forA=41.\left(3.20+...+3^{2009}.20\right)⋮41soA=3+3^3+3^5+3^7+...+3^{2015}⋮41\)
Bài 1:
a) Đặt A = 1 + 7 + 72 + 73 + ... + 72016
7A = 7 + 72 + 73 + 74 + ... + 72017
7A - A = (7 + 72 + 73 + 74 + ... + 72017) - (1 + 7 + 72 + 73 + ... + 72016)
6A = 72017 - 1
\(A=\frac{7^{2017}-1}{6}\)
b) Đặt B = 1 + 4 + 42 + 43 + ... + 42017
4B = 4 + 42 + 43 + 44 + ... + 42018
4B - B = (4 + 42 + 43 + 44 + ... + 42018) - (1 + 4 + 42 + 43 + ... + 42017)
3B = 42018 - 1
\(B=\frac{4^{2018}-1}{3}\)
Bài 2:
a) Ta có: \(14\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}-1⋮13\left(đpcm\right)\)
b) Ta có: \(2015\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}-1⋮2014\left(đpcm\right)\)
Sorry mình thiếu 1+7+72+73+...+72016 câu dưới cũng thiếu 4 nha
A=(2^1+2^2+2^3+2^4+2^5+2^6)+................+(2^2005+2^2006+2^2007+2^2008+2^2009+2^2010)
A=2^1(1+2+2^2+2^3+2^4+2^5)+...................+2^2005(1+2+2^2+2^3+2^4+2^5)
A=2.63+......................+2^2005.63
A=63.(2+..............................+2^2005)
VÌ 63 CHIA HẾT CHO 3 VÀ 7 VẬY A CHIA HẾT CHO 3 VÀ 7.
TICK CHO MÌNH NHA
*Chứng minh A chia hết cho 4
Ta có: \(A=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2015}+3^{2016}\right)\)
\(=3^1.\left(1+3\right)+3^3\left(1+3\right)+...+3^{2015}\left(1+3\right)\)
\(=4\left(3^1+3^3+...+3^{2015}\right)⋮4^{\left(đpcm\right)}\)
*Chứng minh A chia hết cho 13
Ta có: \(A=\left(3^1+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)
\(=3\left(1+3^1+3^2\right)+...+3^{2014}\left(1+3^1+3^2\right)\)
\(=13\left(3+...+3^{2014}\right)⋮13^{\left(đpcm\right)}\)
A = 3 + 32 + 33 + 34 + ... + 32015 + 32016
A = (3 + 32) + (33 + 34) + ... + (32015 + 32016)
A = 3(1 + 3) + 33(1 + 3) + ... + 32015(1 + 3)
A = 3.4 + 33.4 + ... + 32015.4
A = 4(3 + 33 + ... + 32015)
Vì 4(3 + 33 + ... + 32015) \(⋮\) 4 nên A \(⋮\) 4
Vậy A \(⋮\) 4
A = 3 + 32 + 33 + 34 + ... + 32015 + 32016
A = (3 + 32 + 33) + (34 + 35 + 36) + ... + (32014 + 32015 + 32016)
A = 3(1 + 3 + 32) + 34(1 + 3 + 32) + ... + 32014(1 + 3 + 32)
A = 3.13 + 34.13 + ... + 32014.13
A = 13(3 + 34 + ... + 32014)
Vì 13(3 + 34 + ... + 32014) \(⋮\) 13 nên A \(⋮\) 13
Vậy A \(⋮\) 13
A=(21+22+23+24+25+26) + . . . + (22005+22006+22007+22008+22009+22010)
A=2^1(1+2+22+23+24+25)+...................+22005(1+2+22+23+24+25)
A=2.63+......................+22005.63
A=63.(2+..............................+22005)
VÌ 63 CHIA HẾT CHO 3 VÀ 7 VẬY A CHIA HẾT CHO 3 VÀ 7.
\(A=1+3+3^2+3^3+3^4+...+3^{2014}+3^{2015}\)
\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{2013}\left(1+3+3^2\right)\)
\(A=13+3^3.13+...+3^{2013}.13\)
\(A=13\left(1+3^3+...+3^{2013}\right)\) chia hết cho 13
\(A=1+3+3^2+3^3+3^4+...+3^{2014}+3^{2015}\)
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+....+\left(3^{2013}+3^{2014}+3^{2015}\right)\)
\(A=\left(1+3+3^2\right)+3^3\left(1+3+2^2\right)+....+3^{2013}\left(1+3+3^2\right)\)
\(A=13+3^3.13+....+3^{2013}.13\)
\(A=13\left(1+3^3+...+3^{2013}\right)\)chia hết cho 13 (Đpcm)