D = 2^2012 - (1+2+2^2+...+2^2011).
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
19 tháng 2 2017
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(1+\frac{2011}{2}\right)+...+\left(1+\frac{2}{2011}\right)+\left(1+\frac{1}{2012}\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013\)
Vậy x = 2013
IL
0
DL
1
12 tháng 2 2017
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right).x=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)
Đặt \(A=1+2+2^2+...+2^{2011}\)
=>\(2A=2+2^2+...+2^{2012}\)
=>\(2A-A=2+2^2+...+2^{2012}-1-2-...-2^{2011}\)
=>\(A=2^{2012}-1\)
\(D=2^{2012}-A=2^{2012}-2^{2012}+1=1\)
Đặt 𝐴=1+2+22+...+22011A=1+2+22+...+22011
=>2𝐴=2+22+...+220122A=2+22+...+22012
=>2𝐴−𝐴=2+22+...+22012−1−2−...−220112A−A=2+22+...+22012−1−2−...−22011
=>𝐴=22012−1A=22012−1
𝐷=22012−𝐴=22012−22012+1=1D=22012−A=22012−22012+1=1