Question

Thực hiện phép tính :

A = \(\left(1-\dfrac{1}{1+2}\right).\left(1-\dfrac{1}{1+2+3}\right).....\left(1-\dfrac{1}{1+2+3+....+2006}\right)\)

Answer 1

\(A=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)........\left(1-\dfrac{1}{1+2+3+.....+2006}\right)\)

\(=\left(1-\dfrac{1}{\dfrac{\left(1+2\right).2}{2}}\right).\left(1-\dfrac{1}{\dfrac{\left(1+3\right).3}{2}}\right)........+\left(1-\dfrac{1}{\dfrac{\left(1+2006\right).2006}{2}}\right)\)

\(=\left(1-\dfrac{1}{\dfrac{3.2}{2}}\right).\left(1-\dfrac{1}{\dfrac{4.3}{2}}\right)......\left(1-\dfrac{1}{\dfrac{2007.2006}{2}}\right)\)

\(=\left(1-1:\dfrac{3.2}{2}\right).\left(1-1:\dfrac{4.3}{2}\right)........\left(1-1:\dfrac{2007.2006}{2}\right)\)

\(=\left(1-\dfrac{2}{3.2}\right).\left(1-\dfrac{2}{4.3}\right).........\left(1-\dfrac{2}{2007.2006}\right)\)