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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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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Theo đầu bài ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2015}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2015}\)
\(\Rightarrow S=P\)
Vậy ( S - P )2016 = 02016 = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
S - P = (1 - 1/2 + 1/3 -1/4+ ...+ 1/1007 - 1/1008 + ...+ 1/2013 - 1/2014 + 1/2015) - (1/1008 + 1/1009 + ...+1/2014 + 1/2015)
=1 - 1/2 + 1/3 - 1/4 + ... + 1007 -2/1008 - ... - 2/2014
= 1 - 1/2 + 1/3 - 1/4 + ...+ 1/1007 - 2/1008 - 2/1010 - ...- 2/2012 - 2/2014
= 1 - 1/2 + 1/3 - 1/4 + ....+ 1007 - 1/504 - 1/505 - ...- 1/1006 - 1/1007
= 1 - 1/2 + 1/3 - 1/4 + ...1/503 - 1/504 + 1/505 + ...+ 1/1005 - 1/1006 + 1/1007 - 1/504 - 1/505 - ...- 1/1006 - 1/1007
= 1 - 1/2 + 1/3 - 1/4 + ...1/503 - 2/504 - 2/506 - ..- 2/1006
= 1 - 1/2 + 1/3 - 1/4 + ...1/503 - 1/252 - 1/253 - ...- 1/503
Lại tiếp tục như trên, Lẻ mất, chẵn còn => S - P = 0 => (S-P)2015 = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(S=2^{2015}-2^{2014}-...-2-1\)
\(S=2^{2015}-\left(2^{2014}+...+2+1\right)\)
Đặt \(A=2^{2014}+...+2+1\) ta có :
\(2A=2^{2015}+...+2^2+2\)
\(2A-A=\left(2^{2015}+...+2^2+2\right)-\left(2^{2014}+...+2+1\right)\)
\(A=2^{2015}-1\)
\(\Rightarrow\)\(S=2^{2015}-A=2^{2015}-\left(2^{2015}-1\right)=2^{2015}-2^{2015}+1=1\)
Vậy \(S=1\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi A = 2^2015 - 2^2014 - ... -2 - 1
2A = 2^2016 - 2^2015 - ... -2^2 - 2
2A - A = 2^2016 - 2 ^2015 - ...-2^2-2 - 2^2015 +2^2014 + .... +2 + 1
A = 2^2016 - 2.2^2015 + 1
A = 2^2016 - 2^2016 + 1
A = 1
h dung nha