Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016<0,2
\(4.\left(-\frac{1}{2}\right)^3-3.\left(-\frac{1}{2}\right)^2-3.\left(-\frac{1}{2}\right)^1=4.-\frac{1}{8}-3.\frac{1}{4}-3.-\frac{1}{2}\)
\(=-\frac{1}{2}-\frac{3}{4}-\left(-\frac{3}{2}\right)\)
\(=-\frac{1}{2}-\frac{3}{4}+\frac{3}{2}\)
\(=\left(-\frac{1}{2}+\frac{3}{2}\right)-\frac{3}{4}\)
\(=1-\frac{3}{4}\)
\(=\frac{1}{4}\)
Hok tốt nha^^
\(4.\left(-\frac{1}{2}\right)^3-2.\left(-\frac{1}{2}\right)^2-3.\left(-\frac{1}{2}\right)^1\)
\(=4.\left(-\frac{1}{2}.\left(-\frac{1}{2}\right).\left(-\frac{1}{2}\right)\right)-2.\left(-\frac{1}{2}.\left(-\frac{1}{2}\right)\right)-3.\left(-\frac{1}{2}\right)\)( sửa ngoặc vuông giúp mk )
\(=4.\left(-\frac{1}{8}\right)-2.\left(\frac{1}{4}\right)-3.\left(-\frac{1}{2}\right)\)
\(=-\frac{1}{2}-\frac{1}{2}+\frac{3}{2}\)
\(=1+\frac{3}{2}\)
\(=\frac{5}{2}\)
\(4\times\left(-\frac{1}{2}\right)^3-2\times\left(-\frac{1}{2}\right)^2-3\times\left(-\frac{1}{2}\right)^1\)
\(=4\times\left(-\frac{1}{2}\right)^3-2\times\left(-\frac{1}{2}\right)^2-3\times\left(-\frac{1}{2}\right)\)
\(=4\times\left(-\frac{1}{8}\right)-2\times\frac{1}{4}-3\times\left(-\frac{1}{2}\right)\)
\(=-\frac{1}{2}-\frac{1}{2}+\frac{3}{2}\)
\(=1+\frac{3}{2}\)
\(=\frac{2}{2}+\frac{3}{2}\)
\(=\frac{5}{2}\)
\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2015^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\right)\)
\(=1-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\right)>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=1-\left(1-\frac{1}{2015}\right)=1-\frac{2014}{2015}=\frac{1}{2015}\)
=> \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2015^2}>\frac{1}{2015}\left(\text{đpcm}\right)\)