\(=1+1.79^3-1.71^3=1.735128\)

\(=1+1.69\cdot1.3^2+1.65=5.5061\)

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

\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{19}\)(2)

Lấy (2) trừ đi (1) ta có :

\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{20}\)

\(\Rightarrow A=1-\left(\frac{1}{2}\right)^{20}\)

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

          =\(\frac{n.\left(n+1\right)-2}{n\left(n+1\right)}\)

          =\(\frac{n^2+n-2}{n.\left(n+1\right)}\) 

          =\(\frac{\left(n^2-1\right)+\left(n-1\right)}{n.\left(n+1\right)}\)

          =\(\frac{\left(n-1\right).\left(n+1\right)+\left(n-1\right)}{n.\left(n+1\right)}\) 

          =\(\frac{\left(n-1\right).\left(n+1+1\right)}{n.\left(n+1\right)}\)

          =\(\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\)

=>\(1-\frac{2}{n.\left(n+1\right)}=\frac{\left(n-1\right).\left(n+2\right)}{n.\left(n+1\right)}\left(1\right)\)

Lại có: \(M=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)....\left(1-\frac{2}{99.100}\right)\)

=>      \(M=\left(1-\frac{2}{2.\left(2+1\right)}\right).\left(1-\frac{2}{3.\left(3+1\right)}\right).\left(1-\frac{2}{4.\left(4+1\right)}\right)....\left(1-\frac{2}{99.\left(99+1\right)}\right)\left(2\right)\)

Thay (1) vào (2) ta được:

         \(M=\frac{\left(2-1\right).\left(2+2\right)}{2.\left(2+1\right)}.\frac{\left(3-1\right).\left(3+2\right)}{3.\left(3+1\right)}.\frac{\left(4-1\right).\left(4+2\right)}{4.\left(4+1\right)}...\frac{\left(99-1\right).\left(99+2\right)}{99.\left(99+1\right)}\)

=>     \(M=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{98.101}{99.100}\)

=>     \(M=\frac{1.4.2.5.3.6....98.101}{2.3.3.4.4.5....99.100}\)

=>     \(M=\frac{\left(1.2.3....98\right).\left(4.5.6....101\right)}{\left(2.3.4....99\right).\left(3.4.5....100\right)}\)

=>     \(M=\frac{1.101}{99.3}\)

=> \(M=\frac{101}{297}\)

Vậy \(M=\frac{101}{297}\)

\(\frac{1}{2}S=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\)

\(\Rightarrow\left(\frac{1}{2}S\right)-S=\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\right)-\left(\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{20}\right)\)

\(\Rightarrow-\frac{1}{2}S=\left(\frac{1}{2}\right)^{21}-\left(\frac{1}{2}\right)\)

\(\Rightarrow S=\frac{\left(\left(\frac{1}{2}\right)^{21}-\frac{1}{2}\right)}{-\frac{1}{2}}\)