M= 2+ 2^2+ 2^3+ 2^4+....+ 2^2017+ 2^2018
tính M
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)đề \(\Rightarrow2M=2^2+2^3+2^4+...+2^{2019}
\Rightarrow M=2^{2019}-2\)
b)đề \(\Rightarrow M=(2+2^2)+(2^3+2^4)+...+(2^{2017}+2^{2018})\)
\(\Rightarrow M=2.3+3.\left(2^3\right)+3.2^4+...+3.2^{2017}\)
\(\Rightarrow M⋮3\left(đpcm\right)\)
Ta có : 2M = 2 +\(\frac{3}{2}\)+\(\frac{4}{2^2}\)+...+\(\frac{2017}{2^{2015}}\)+ \(\frac{2018}{2^{2016}}\)
2M - M = 2 + \(\frac{3}{2}\)- \(\frac{2}{2}\)+ \(\frac{4}{2^2}\)-\(\frac{3}{2^2}\)+...+\(\frac{2017}{2^{2015}}\)-\(\frac{2016}{2^{2015}}\)+ \(\frac{2018}{2^{2016}}\)-\(\frac{2017}{2^{2016}}\)-\(\frac{2018}{2^{2017}}\)
M = 2 + \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{2015}}\)+ \(\frac{1}{2^{2016}}\)-\(\frac{2018}{2^{2017}}\)
Đặt N = \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{2016}}\)
Ta có :2N = 1 + \(\frac{1}{2}\)+\(\frac{1}{2^2}\)+ .....+\(\frac{1}{2^{2015}}\)
2N - N = 1\(\frac{1}{2^{2016}}\)
Vậy N < 1
Nên M < 2 + 1 - \(\frac{2018}{2^{2017}}\)= 3 -\(\frac{2018}{2^{2017}}\)
Vậy M < 3
Ta có : M = 2 + 22 + 23 + 24 + .... + 22017 + 22018
=> 2M = 22 + 23 + 24 + 25 + .... + 22018 + 22019
=> 2M - M = ( 22 + 23 + 24 + 25 + .... + 22018 + 22019 ) - (2 + 22 + 23 + 24 + .... + 22017 + 22018 )
=> M = 22019 - 2
b) Lại có M = 2 + 22 + 23 + 24 + .... + 22017 + 22018
= (2 + 22) + (23 + 24) + .... + (22017 + 22018)
= 2(2 + 1) + 23(2 + 1) + ... + 22017(2 + 1)
= (2 + 1)(2 + 23 + .... + 22017)
= 3(2 + 23 + .... + 22017)
=> M \(⋮\)3 (ĐPCM)
\(a,\)\(M=2+2^2+2^3+2^4+...+2^{2017}+2^{2018}\)
\(2M=2^2+2^3+2^4+2^5+....+2^{2018}+2^{2019}\)
\(M=2^{2019}-2\)
\(b,\)\(M=2+2^2+2^3+2^4+....+2^{2017}+2^{2018}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^{2017}+2^{2018}\right)\)
\(=2\left(2+1\right)+2^3\left(2+1\right)+....+2^{2017}\left(2+1\right)\)
\(=3\left(2+2^3+...+2^{2017}\right)⋮3\)
2M=2^2+2^3+...+2^2019
=>2M-M=2^2+2^3+...+2^2019-2-2^2-...-2^2018
=>M=2^2019-2