Evaluate:  ( $\frac{1}{\sqrt{16}-\sqrt{15}})-(\frac{1}{\sqrt{15}-\sqrt{14}})+(\frac{1}{\sqrt{14}-\sqrt{13}})-(\frac{1}{\sqrt{13}-\sqrt{12}})+(\frac{1}{\sqrt{12}-\sqrt{11}})\; =?$

Correct Option / Answer:
The correct answer is $4+\sqrt{11}$.

Step-by-Step Explanation:

Step 1: Rationalize each term in the expression
To simplify a fraction of the form $\frac{1}{\sqrt{n}-\sqrt{n-1}}$, we rationalize the denominator by multiplying both the numerator and the denominator by its conjugate, $\sqrt{n}+\sqrt{n-1}$:

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

Using the algebraic identity $(a-b)(a+b)=a^2-b^2$, the denominator becomes:

${\left(\sqrt{n}\right)}^2-{\left(\sqrt{n-1}\right)}^2=n-(n-1)=1$

Therefore, each term simplifies to:

$\frac{1}{\sqrt{n}-\sqrt{n-1}}=\sqrt{n}+\sqrt{n-1}$

Step 2: Simplify each term individually

1. $\frac{1}{\sqrt{16}-\sqrt{15}}=\sqrt{16}+\sqrt{15}$
2. $\frac{1}{\sqrt{15}-\sqrt{14}}=\sqrt{15}+\sqrt{14}$
3. $\frac{1}{\sqrt{14}-\sqrt{13}}=\sqrt{14}+\sqrt{13}$
4. $\frac{1}{\sqrt{13}-\sqrt{12}}=\sqrt{13}+\sqrt{12}$
5. $\frac{1}{\sqrt{12}-\sqrt{11}}=\sqrt{12}+\sqrt{11}$

Step 3: Substitute the rationalized terms back into the original expression

$(\sqrt{16}+\sqrt{15})-(\sqrt{15}+\sqrt{14})+(\sqrt{14}+\sqrt{13})-(\sqrt{13}+\sqrt{12})+(\sqrt{12}+\sqrt{11})$

Step 4: Expand the brackets and cancel out intermediate terms

$=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+\sqrt{14}+\sqrt{13}-\sqrt{13}-\sqrt{12}+\sqrt{12}+\sqrt{11}$

Notice that all intermediate terms ($\sqrt{15}$, $\sqrt{14}$, $\sqrt{13}$, and $\sqrt{12}$) cancel out:

$=\sqrt{16}+\sqrt{11}$

Since $\sqrt{16}=4$, we get:

$=4+\sqrt{11}$

Thus, the value of the given expression is $4+\sqrt{11}$.