Từ công thức:\(1+2+........+n=\frac{n.\left(n+1\right)}{2}\)

Cho \(n\in\)N*.CMR:\(\frac{1}{n}.\left(1+2+...+n\right)=\frac{n+1}{2}\)

Ta có:\(\frac{1}{n}.\left(1+2+......+n\right)=\frac{1}{n}.\frac{n\left(n+1\right)}{2}=\frac{n+1}{2}\)

Ta có:\(1+\frac{1}{2}\left(1+2\right)+......+\frac{1}{20}.\left(1+2+.....+20\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+........+\frac{1}{20}.\frac{20\left(20+1\right)}{2}\)

\(=1+\frac{3}{2}+...............+\frac{21}{2}\)

\(=\frac{2+3+......+21}{2}\)

\(=\frac{230}{2}=165\)

\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)

\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)

Ta có : 

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(1+1-\frac{2}{x+1}=\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=2-\frac{2003}{2005}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=2:\frac{2007}{2005}\)

\(\Leftrightarrow\)\(x+1=\frac{4010}{2007}\)

\(\Leftrightarrow\)\(x=\frac{4010}{2007}-1\)

\(\Leftrightarrow\)\(x=\frac{2003}{2007}\)

Vậy \(x=\frac{2003}{2007}\)

Chúc bạn học tốt ~ 

A=(3+1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

=>2A=2.(3+1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

=(3-1)(3+1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

=(3²-1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

=(3⁴+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

=(3⁸-1)(3⁸+1)(3¹⁶+1)

=(3¹⁶-1)(3¹⁶+1)

=3³²-1

=>A=\(\frac{3^{32}-1}{2}