Question

How to do this question?

Answer 1

Option D

Explanation:

Take any pair:

`1/(sqrtn-1) - 1/(sqrtn+1)`

`= (sqrtn + 1 - sqrtn +1)/((sqrtn-1)(sqrtn+1))`

`= (2 )/(n-1)`

So, term by term:

`{(n = 2, 2/1),(n=3, 2/2), (...,...),(n=20, 2/19) :}`

So you have:

`2/1 + 2/2 + ... + 2/19 = sum_(n=1)^19 2/n`

Answer 2

Option D. See process below

Explanation:

We notice that

`1/(sqrt2-1)=(sqrt2+1)/(2-1)=(sqrt2+1)/1`
`1/(sqrt2+1)=(sqrt2-1)/(2-1)=(sqrt2-1)/1`

The sum of both expresions (the two first sum terms) is 2

`1/(sqrt3-1)=(sqrt3+1)/(3-1)=(sqrt3+1)/2`
`1/(sqrt3+1)=(sqrt3-1)/(3-1)=(sqrt3-1)/2`

The sum of both expresions (third and fourth) is 1

`1/(sqrt4-1)=(sqrt4+1)/(4-1)=(sqrt4+1)/3`
`1/(sqrt4+1)=(sqrt4-1)/(4-1)=(sqrt4-1)/3`

The sum of both terms is `2/3`

For last two terms

`1/(sqrt20-1)=(sqrt20+1)/(20-1)=(sqrt20+1)/19`
`1/(sqrt20+1)=(sqrt20-1)/(20-1)=(sqrt20-1)/19`

The sum of both terms is `2/19`

So we have

`2+1+2/3+...+2/19=sum_(n=1)^(n=19)2/n`

Correct answer: D