Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=1+22+23+....+22016+22017
=>2A=2+22+23+24+.....+22016+22017+22018
=>2A-A=(2+22+23+24+....+22016+22017+22018) - (1+2+22+23+....+22016+22017)
=>A=22018-1
Tìm số dư khi chia A cho 7 biết rằng : A = 1 + 2 + 22 + 23 + ... + 22016 + 22017
A=1+22+23+....+22016+22017
=>2A=2+22+23+24+.....+22016+22017+22018
=>2A-A=(2+22+23+24+....+22016+22017+22018) - (1+2+22+23+....+22016+22017)
=>A=22018-1
Ta Có :
A =20+21+22+23+.....+22016
=>A = 20(1+2+4) + 23(1+2+4) + ...... + 22014(1+2+4)
=>A = 1.7 + 23.7 + ....... +22014 . 7
=>A = 7.(1+23+......+22014)
=> A chia hết cho 7 (ĐPCM)
Đúng thì cho tích nha bạn !
\(A=2^0+2^1+2^2+...+2^{2016}\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2A-A=2^{2017}-2^0\)
\(\Rightarrow A=2^{2017}-1\)
ma chia cho may ? ko biet chia cho so nao lam sao tim so du
\(A=1+3+3^2+3^3+3^4+3^5+.....+3^{2017}\)
\(=1+3+\left(3^2+3^3+3^4+3^5\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}+3^{2017}\right)\)
\(=4+3^2\left(1+3+3^2+3^3\right)+.....+3^{2014}\left(1+3+3^2+3^3\right)\)
\(=4+3^2\cdot40+....+3^{2014}\cdot40\)
\(=4+40\left(3^2+.....+3^{2014}\right)\) chia 40 dư 4.
\(\frac{3-x}{2016}-1=\frac{2-x}{2017}+\frac{1-x}{2018}\)
\(\Rightarrow\frac{3-x}{2016}-1+2=\frac{2-x}{2017}+\frac{1-x}{2018}+2\)(thêm 2 vô mỗi vế)
\(\Rightarrow\frac{3-x}{2016}+1=\left(\frac{2-x}{2017}+1\right)+\left(\frac{1-x}{2018}+1\right)\)
\(\Rightarrow\frac{2019-x}{2016}=\frac{2019-x}{2017}+\frac{2019-x}{2018}\)
\(\Rightarrow\left(2019-x\right)\cdot\frac{1}{2016}=\left(2019-x\right)\left(\frac{1}{2017}+\frac{1}{2018}\right)\)
\(\Rightarrow2019-x=0\)
\(\Rightarrow x=2019\)
ta có : \(A=1+5+5^2+...+5^{2016}+5^{2017}\) có \(2017\) số hạng
mà \(2017\) chia cho \(3\) dư \(1\)
\(\Rightarrow\) ta có thể gợp \(A\) lại từng tổng số hạng như sau
\(A=1+\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{2015}+5^{2016}+5^{2017}\right)\)
\(A=1+5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{2015}\left(1+5+5^2\right)\)
\(A=1+5\left(1+5+25\right)+5^4\left(1+5+25\right)+...+5^{2015}\left(1+5+25\right)\)
\(A=1+5.31+5^4.31+...+5^{2015}.31\)
\(A=1+31\left(5+5^4+...+5^{2015}\right)\)
ta có : \(31\left(5+5^4=...+5^{2015}\right)⋮31\)
\(\Rightarrow31\left(5+5^4+...+5^{2015}\right)\) chia hết cho \(31\)
mà \(1< 3\) nên \(1\) không thể chia cho 3 thêm được nữa
\(\Rightarrow A=1+31\left(5+5^4+...+5^{2015}\right)\) chia cho \(31\) dư \(1\)
vậy \(A=1+31\left(5+5^4+...+5^{2015}\right)\) chia cho \(31\) dư \(1\)