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Ta có :
\(S=2^{2015}-2^{2014}-..............-2-1\)
\(\Leftrightarrow S=2^{2015}-\left(2^{2014}+2^{2013}+...........+2+1\right)\)
Đặt :
\(A=2^{2014}+2^{2013}+.........+2+1\)
\(\Leftrightarrow2A=2^{2015}+2^{2014}+.............+2\)
\(\Leftrightarrow2A-A=\left(2^{2015}+2^{2014}+..........+2\right)-\left(2^{2014}+2^{2013}+.........+1\right)\)
\(\Leftrightarrow A=2^{2015}-1\)
\(\Leftrightarrow S=2^{2015}-\left(2^{2015}+1\right)\)
\(\Leftrightarrow S=2^{2015}-2^{2015}+1\)
\(\Leftrightarrow S=0+1=1\)
\(S=2^{2015}-2^{2014}-2^{2013}-...2-1\)
\(2S=2^{2015}-2^{2014}-2^{2013}-...-2\)
\(2S-S=2^{2015}-2^{2014}-2^{2014}-2^{2013}+2^{2013}-...-2+2+1\)
\(S=2^{2015}-2.2^{2014}+1\)
\(S=2^{2015}-2^{2015}+1=1\)
Tham khảo, chúc bạn học giỏi! Haizzz
Đặt \(B=2^{2014}+2^{2013}+...+2+1\)
\(\Leftrightarrow2B=2^{2015}+2^{2014}+...+2^2+2\)
\(\Leftrightarrow B=2^{2015}-1\)
\(A=2^{2015}-B=2^{2015}-2^{2015}+1=1\)
Ta có: A = 22015 - 22014 - 22013 - ... - 2 - 1
nên 2A = 22016 - 22015 - 22014 - ... - 22 - 2
2A - A = (22016 - 22015 - 22014 - ... - 22 - 2) - (22015 - 22014 - 22013 - ... - 2 - 1)
A = 22016 - 2.22015 + 1
A = 22016 - 22016 + 1 = 1
Vậy, 2015A = 20151 = 2015
Đặt A = 22015 + 22014 + 22013 + ... + 22 + 2 + 1
=> 2A = 2 + 22 + 23 + ...... + 22015 + 22016
=> 2A - A = 22016 - 1
=> A = 22016 - 1
cái này là lớp 6 mà
đặt tên biểu thức trên là A
Ta có :
\(A=2^{2015}+2^{2014}+2^{2013}+...+2^2+2^2+1\)
\(2A=2.\left(2^{2015}+2^{2014}+2^{2013}+...+2^2+2+1\right)\)
\(2A=2^{2016}+2^{2015}+2^{2014}+...+2^3+2^2+2\)
\(2A-A=\left(2^{2016}+2^{2015}+2^{2014}+...+2^3+2^2+2\right)-\left(2^{2015}+2^{2014}+2^{2013}+...+2^2+2+1\right)\)\(A=2^{2016}-1\)
Nguyễn Quang Trung CTV làm sơ ý quá
a) Đặt \(A=2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2-2^1\)
\(\Rightarrow2A=2^{2017}-2^{2016}+2^{2015}-2^{2014}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{2017}-2^{2015}+2^{2014}-2^{2013}+...+2^3-2^2\right)+\left(2^{2016}-2^{2015}+2^{2014}-2^{2013}+...+2^2+2^1\right)\)
\(\Rightarrow3A=2^{2017}+1\)
\(\Rightarrow A=\frac{2^{2017}+1}{3}\)
b) Đặt \(B=3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\)
\(\Rightarrow3B=3^{1001}-3^{1000}+3^{999}-3^{997}+...+3^3-3^2+3^1\)
\(\Rightarrow3B+B=\left(3^{1001}-3^{1000}+3^{999}-3^{998}+...+3^3-3^2+3^1\right)+\left(3^{1000}-3^{999}+3^{998}-3^{997}+...+3^2-3^1+3^0\right)\)
\(\Rightarrow4B=3^{1001}+3^0\)
\(\Rightarrow B=\frac{3^{1001}+1}{4}\)
a) Đặt A = 22016 - 22015 + 22014 - 22013 + ... + 22 - 21
2A = 22017 - 22016 + 22015 - 22014 + ... + 23 - 22
2A + A = (22017 - 22016 + 22015 - 22014 + ... + 23 - 22) + (22016 - 22015 + 22014 - 22013 + ... + 22 - 21)
3A = 22017 - 21
3A = 22017 - 2
\(A=\frac{2^{2017}-2}{3}\)
b) lm tương tự câu a
2 S = 22016 - ( 22015 + 2 2014 + 22013 +.....+ 23 + 2 2 + 2 )
2S - S = 2 2016 + 1
S = 22015- 22014- 22013-.......-22-21-20
2S = 22016 - 22015 -22014 - 22013 -..........- 23 -22 -21
2S -S = 22016 -22015 -22014 -22013 -....- 23-22 -21 - 22015 + 22014 + 22013 +.....+ 23 +22+21+20
= 22016 - 2x22015 + 20
= 20=1