Question

Could you tell me the answer of 3(b)? Pleaseeeeee

Answer 1

(b)

`\displaystyle (2005)/(2006)`

Explanation:

(a)

First, we have to prove that:

`\displaystyle (1)/(x)-(1)/(x+1)=(1)/(x(x+1))`

Operate:

`\displaystyle (1)/(x)-(1)/(x+1)=(x+1-x)/(x(x+1))`

Simplifying:

`\displaystyle (1)/(x)-(1)/(x+1)=(1)/(x(x+1)) [1]`

(b)

Now we use the above expression to find the sum:

`\displaystyle S=(1)/(1* 2)+(1)/(2* 3)+(1)/(3* 4)+...+(1)/(2005* 2006)`

Let's use [1] for each term of the sum:

`\displaystyle (1)/(1* 2)=(1)/(1)-(1)/(2)`

`\displaystyle (1)/(2* 2)=(1)/(2)-(1)/(3)`

`\displaystyle (1)/(3* 4)=(1)/(3)-(1)/(4)`

`\displaystyle (1)/(2005* 2006)=(1)/(2005)-(1)/(2006)`

Substituting:

`\displaystyle S=(1)/(1)-(1)/(2)+(1)/(2)-(1)/(3)+(1)/(3)-(1)/(4)+....+(1)/(2005)-(1)/(2006)`

Note the terms 1/2 and -1/2, 1/3 and -1/3, etc. are canceled out, leaving only the first and last terms:

`\displaystyle S=(1)/(1)-(1)/(2006)=(2006-1)/(2006)`

`\boxed{\displaystyle S=(2005)/(2006)}`