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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 = 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) \(⋮4\)
Chia đề bài thành 2 phần như sau:
Phần thứ nhất: Chứng tỏ B chia hết cho 4. Ta có:
\(B=3+3^2+3^3+3^4+3^5+...+3^{2015}+3^{2016}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{2015}+3^{2016}\right)\)
\(B=\left(3\cdot1+3.3\right)+\left(3^3\cdot1+3^3\cdot3\right)+\left(3^5\cdot1+3^5\cdot3\right)+...+\left(3^{2015}\cdot1+3^{2015}\cdot3\right)\)
\(B=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+...+3^{2015}\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{2015}\cdot4\)
\(B=4\left(3+3^3+3^5+...+3^{2015}\right)\)
Do B có một thừa số là 4 nên B chia hết cho 4. Đã chứng minh được phần thứ nhất.
Phần thứ hai: Chứng tỏ B chia hết cho 13. Ta có:
\(B=3+3^2+3^3+3^4+3^5+...+3^{2015}+3^{2016}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)
\(B=\left(3\cdot1+3\cdot3+3\cdot9\right)+\left(3^4\cdot1+3^4\cdot3+3^4\cdot9\right)+...+\left(3^{2014}\cdot1+3^{2014}\cdot3+3^{2014}\cdot9\right)\)
\(B=3\left(1+3+9\right)+3^4\left(1+3+9\right)+...+3^{2014}\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+...+3^{2014}\cdot13\)
\(B=13\left(3+3^4+...+3^{2014}\right)\)
Do B có thừa số 13 nên B chia hết cho 13. Phần thứ hai đã được chứng minh.
Qua hai phần trên, ta kết luận: B chia hết cho 4 và 13.
B = 3+3^2+3^3+3^4+..+3^2015+3^2016
=>B=(3+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^2015+3^2016)
=>B=12+3^2(3+3^2)+3^4+(3+3^2)+...+3^2014(3+3^2)
=>B=12+3^2.12+3^4.12+...+3^2014.12
=>B=12(1+3^2+3^4+...+3^2014)
=>?B=4.3.(1+3^2+3^4+...+3^2014)=>B chia hết cho 4
B=3+3^2+3^3+3^4+...+3^2015+3^2016
=>B=(3+3^2+3^3)+(3^4+3^5+3^6)+(3^7+3^8+3^9)+...+(3^2014+3^2015+3^2016)
=>B=39+3^3(3+3^2+3^3)+3^3(3+3^2+3^3)+3^6(3+3^2+3^3)+...+3^2013(3+3^2+3^3)
=>B=39+3^3.39+3^6.39+...+3^2013.39
=>B=39(1+3^3+3^6+...+3^2013)
=>b=13.3.(1+3^3+3^6+....+3^2013)=>B chia hết cho 13
8 Cho A = 32016 + 32015 + ... + 32 + 3
a) Chứng minh A chia hết cho 4
b) Chứng minh A chia hết cho 13
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
\(A=1+3+3^2+3^3+3^4+...+3^{2014}+3^{2015}\)
\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{2013}\left(1+3+3^2\right)\)
\(A=13+3^3.13+...+3^{2013}.13\)
\(A=13\left(1+3^3+...+3^{2013}\right)\) 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)
Mẫu câu a)!! những câu khác ko lm đc ib!
a) Ta có:
\(A=2+2^2+2^3+2^4+...+2^{2010}.\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{2009}.3\)
\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)
Ta có:
\(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{2008}.7\)
\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)
b,\(B=3+3^2+3^3+3^4+...+3^{2010}.\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{2009}.4\)
\(=4.\left(3+3^3+...+3^{2009}\right)⋮4\)
\(B=3+3^2+3^3+3^4+...+3^{2010}\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{2008}.13\)
\(=13\left(3+3^4+...+3^{2008}\right)⋮13\)
*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: \(A=\left(3^1+3^2+3^3\right)+...+\left(3^{2014}+3^{2015}+3^{2016}\right)\)
\(=3\left(1+3^1+3^2\right)+...+3^{2014}\left(1+3^1+3^2\right)\)
\(=13\left(3+...+3^{2014}\right)⋮13^{\left(đpcm\right)}\)