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\(C=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+..+\frac{1}{2016}.\left(1+2+3+...+2016\right)\)
\(C=1+\frac{1}{2}.\left(1+2\right).2:2+\frac{1}{3}.\left(1+3\right).3:2+\frac{1}{4}.\left(1+4\right).4:2+...+\frac{1}{2016}.\left(1+2016\right).2016:2\)
\(C=1+3:2+4:2+5:2+...+2017:2\)
\(C=2.\frac{1}{2}+3.\frac{1}{2}+4.\frac{1}{2}+5.\frac{1}{2}+...+2017.\frac{1}{2}\)
\(C=\frac{1}{2}.\left(2+3+4+5+...+2017\right)\)
\(C=\frac{1}{2}.\left(2+2017\right).2016:2\)
\(C=\frac{1}{2}.2019.2016.\frac{1}{2}\)
\(C=2019.504=1017576\)
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
\(A=(1-\frac{1}{1+2})(1-\frac{1}{1+2+3})(1-\frac{1}{1+2+3+4})...(1-\frac{1}{1+2+3+...+2006})\)
\(A=(1-\frac{1}{3})(1-\frac{1}{6})(1-\frac{1}{10})...(1-\frac{1}{2013021})\)
\(A=\frac{2}{3}\cdot\frac{5}{6}\cdot\frac{9}{10}....\frac{2013020}{2013021}\)
Sorry bạn máy tính mình có chút vấn đề để mk làm tiếp :
\(A=\frac{4}{6}\cdot\frac{10}{12}\cdot\frac{18}{20}....\cdot\frac{4026040}{4026042}\)
\(A=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\frac{3\cdot6}{4\cdot5}\cdot...\cdot\frac{2005\cdot2008}{2006\cdot2007}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2005}{2\cdot3\cdot4\cdot...\cdot2006}\cdot\frac{4\cdot5\cdot6\cdot...\cdot2008}{3\cdot4\cdot5\cdot...\cdot2007}\)
\(A=\frac{1}{2006}\cdot\frac{2008}{3}=\frac{1004}{3009}\)
P/S : Hoq chắc :>