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Answer 1 · 2018-06-27T09:33:23

Option D

Take any pair:

$\frac{1}{\sqrt{n} - 1} - \frac{1}{\sqrt{n} + 1}$ $1/(sqrtn-1) - 1/(sqrtn+1)$

$= \frac{\sqrt{n} + 1 - \sqrt{n} + 1}{(\sqrt{n} - 1) (\sqrt{n} + 1)}$ $= (sqrtn + 1 - sqrtn +1)/((sqrtn-1)(sqrtn+1))$

$= \frac{2}{n - 1}$ $= (2 )/(n-1)$

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 · 2018-06-27T09:43:27

Option D. See process below

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