chứng tỏ rằng các tổng sau:
1.( 3 + 32+33+34+35+36+.........+321) chia hết cho 13
2 (52012 + 52011+ 52010) chia hết cho 31
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HV
0
MH
28 tháng 12 2021
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
29 tháng 12 2021
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
N
2
18 tháng 12 2021
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
HH
1
19 tháng 12 2021
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
MQ
1
23 tháng 12 2021
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
BG
2
TH
Thầy Hùng Olm
Manager
VIP
22 tháng 12 2022
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
12 tháng 12 2021
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
12 tháng 12 2021
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
NL
5
T
29 tháng 10 2023
\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\\=(3+3^2)+(3^3+3^4)+(3^5+3^6)+(3^7+3^8)\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+3^7\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+3^7\cdot4\\=4\cdot(3+3^3+3^5+3^7)\)
Vì \(4\cdot(3+3^3+3^5+3^7) \vdots 4\)
nên \(B\vdots4\).
29 tháng 10 2023
`#3107.101107`
\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+\left(3^7+3^8\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+3^7\left(1+3\right)\)
\(=\left(1+3\right)\left(3+3^3+3^5+3^7\right)\)
\(=4\left(3+3^3+3^5+3^7\right)\)
Vì \(4\left(3^3+3^5+3^7\right)\) $\vdots 4$
`\Rightarrow B \vdots 4`
Vậy, `B \vdots 4.`
TT
1
NN
0