Answer:
(b)
[tex]\displaystyle \frac{2005}{2006}[/tex]
Step-by-step explanation:
(a)
First, we have to prove that:
[tex]\displaystyle \frac{1}{x}-\frac{1}{x+1}=\frac{1}{x(x+1)}[/tex]
Operate:
[tex]\displaystyle \frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x(x+1)}[/tex]
Simplifying:
[tex]\displaystyle \frac{1}{x}-\frac{1}{x+1}=\frac{1}{x(x+1)} [1][/tex]
(b)
Now we use the above expression to find the sum:
[tex]\displaystyle S=\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{2005\times 2006}[/tex]
Let's use [1] for each term of the sum:
[tex]\displaystyle \frac{1}{1\times 2}=\frac{1}{1}-\frac{1}{2}[/tex]
[tex]\displaystyle \frac{1}{2\times 2}=\frac{1}{2}-\frac{1}{3}[/tex]
[tex]\displaystyle \frac{1}{3\times 4}=\frac{1}{3}-\frac{1}{4}[/tex]
[tex]\displaystyle \frac{1}{2005\times 2006}=\frac{1}{2005}-\frac{1}{2006}[/tex]
Substituting:
[tex]\displaystyle S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2005}-\frac{1}{2006}[/tex]
Note the terms 1/2 and -1/2, 1/3 and -1/3, etc. are canceled out, leaving only the first and last terms:
[tex]\displaystyle S=\frac{1}{1}-\frac{1}{2006}=\frac{2006-1}{2006}[/tex]
[tex]\boxed{\displaystyle S=\frac{2005}{2006}}[/tex]
