b: Ta có công thức tổng quát sau:
\(1-\frac{1}{1+2+\cdots+n}\)
\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)
\(=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}\)
\(=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)
\(C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+\cdots+2016}\right)\)
\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(2016+2\right)\left(2016-1\right)}{2016\cdot\left(2016+1\right)}\)
\(=\frac{4\cdot5\cdot\ldots\cdot2018}{3\cdot4\cdot\ldots\cdot2017}\cdot\frac{1\cdot2\cdot\ldots\cdot2015}{2\cdot3\cdot\ldots\cdot2016}=\frac{2018}{3}\cdot\frac{1}{2016}=\frac{1009}{1008\cdot3}=\frac{1009}{3024}\)
