Answer:
1003
Step-by-step explanation:
The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.
The simplest form of a telescoping product[tex]a_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}[/tex], in which the products of n terms is
[tex]a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}.[/tex].
In this particular case, [tex]t_{1} \ = \ 2[/tex] , [tex]t_{2} \ = \ 3[/tex], [tex]t_{3} \ = \ 4[/tex], ..... , in which each term follows a recursive formula of [tex]t_{n+1} \ = \ t_{n} \ + \ 1[/tex]. Therefore,
[tex]\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003[/tex]
