Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
... tìm số dư khi chia hết???
nếu nó chia hết thì số dư bằng 0 rồi
S = 2 + 22 + 23 + ...+2100
S \(\times\) 2 = 22 + 23 +...+2100+2101
2S - S = 2101 - 2
S = 2101 - 2
Ta có: ( Sửa đề )
\(A=4+4^2+4^3+...+4^{2021}+4^{2022}\)
\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)
\(A=20+4^2.\left(4+4^2\right)+...+4^{2020}.\left(4+4^2\right)\)
\(A=20+4^2.20+...+4^{2020}.20\)
\(A=20.\left(1+4^2+...+4^{2020}\right)\)
Vì \(20⋮20\) nên \(20.\left(1+4^2+...+4^{2020}\right)\)
Vậy \(A⋮20\)
\(#WendyDang\)
Úi gời cơi cộng chấm chấm chấm :)))
+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=2.3+2^3.3+...+2^{2009}.3\)
\(A=3\left(2+2^3+...+2^{2010}\right)⋮3\)
-> Đpcm
+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{2008}\left(1+2+2^2\right)\)
\(A=2.7+2^4.7+...+2^{2008}.7\)
\(A=7\left(2+2^4+...+2^{2008}\right)⋮7\)
-> Đpcm
\(A=2^1+2^2+...+2^{2010}\)
\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2+2^2+2^3+...+2^{2010}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)
A=2\(^1\)+2\(^2\)+...+2\(^{2010}\)
=(2\(^1\)+2\(^2\))+(2\(^3\)+2\(^4\))+...+(2\(^{2009}\)+2\(^{2010}\))
=2(1+2)+2\(^3\)(1+2)+...+2\(^{2009}\)(1+2)
=3(2+2\(^3\)+...+2\(^{2009}\))⋮3
S = 1 + 2 + 22 + 23 +......+22022
2S = 2 + 22 + 23+........+22022 + 22023
2S- S = 22023 - 1
S = 22023 - 1