What is the value of $\dfrac{1}{{1 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 4 }}........$ up to $15$ terms?
$A) 1$
$B) 2$
$C) 3$
$D) 4$

Hint: Here we have to find the value of the given series.
Then applying rationalization in the first terms of the given series, we will find the value of the given series.
That is rationalizing the denominator to get a ‘telescoping sum’: a sum of terms in which many pairs add up to zero.
Finally we get the required answer.

Formula used: The general pattern of the series is $\dfrac{1}{{\sqrt n + \sqrt {n + 1} }}$
$(a - b)(a + b) = ({a^2} - {b^2})$

Complete step-by-step solution:
It is given the series being $\dfrac{1}{{1 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 4 }}........$ up to $15$ terms.
Here we have to find the value of the given series.
Let us consider the given series is $\dfrac{1}{{1 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 4 }}............\dfrac{1}{{\sqrt {15} + \sqrt {16} }}$
Here $\dfrac{1}{{\sqrt {15} + \sqrt {16} }}$ be the $15^{th}$ term of the series
By using the pattern $\dfrac{1}{{\sqrt n + \sqrt {n + 1} }}$ in the $15^{th}$ term of the series as follows
Now rationalize the first term of the given series,
$ \Rightarrow \dfrac{1}{{1 + \sqrt 2 }} = \dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }}$
Now we are doing multiplying numerator by numerator and denominator by denominator
Hence we get,
$ \Rightarrow \dfrac{{1 - \sqrt 2 }}{{\left( {1 + \sqrt 2 } \right)\left( {1 - \sqrt 2 } \right)}}$
(By using basic identity of algebra $(a - b)(a + b) = ({a^2} - {b^2})$)
$ \Rightarrow \dfrac{{1 - \sqrt 2 }}{{{1^2} - {{\left( {\sqrt 2 } \right)}^2}}}$
On cancel the square root on the denominator term we get,
$ \Rightarrow \dfrac{{1 - \sqrt 2 }}{{1 - 2}}$
We just subtracted the denominator terms,
$ \Rightarrow \dfrac{{1 - \sqrt 2 }}{{ - 1}}$
On rewriting the terms we get,
$ = - \left( {1 - \sqrt 2 } \right)$
Therefore, $\dfrac{1}{{1 + \sqrt 2 }} = - \left( {1 - \sqrt 2 } \right)$
We can rationalize like this for all the terms
Therefore, we get series like this
$
   - (1 - \sqrt 2 ) - (\sqrt 2 - \sqrt 3 ) - (\sqrt 3 - \sqrt 4 ) - (\sqrt 4 - \sqrt 5 ) - (\sqrt 5 - \sqrt 6 ) - (\sqrt 6 - \sqrt 7 ) - (\sqrt 7 - \sqrt 8 ) - (\sqrt 8 - \sqrt 9 ) \\
   - (\sqrt 9 - \sqrt {10} ) - (\sqrt {10} - \sqrt {11} ) - (\sqrt {11} - \sqrt {12} ) - (\sqrt {12} - \sqrt {13} ) - (\sqrt {13} - \sqrt {14} ) - (\sqrt {14} - \sqrt {15} ) - (\sqrt {15} - \sqrt {16} )
 $
We are cancelling the opposite terms in the above expressions we get,
$ \Rightarrow - 1 + 4$
On adding the terms
$ \Rightarrow 3$
$\therefore $ The value for the given series is $3$

Thus the correct option is $\left( C \right)$ that is $3$

Note: Why is there a need for rationalization? The answer is as you can see that after rationalization meaning multiplying and dividing the whole expression by a conjugate, the denominator of the expression is reduced to $1$ by using the basic algebraic properties.
So rationalization will simplify the denominator in such a way that it contains only rational numbers.
So, in a calculation if you find the denominator can be rationalized then go for it, as it will reduce the complexity of the problem.