Evaluate the value of the expression:

$$\left( \frac{1}{\sqrt{16} - \sqrt{15}} \right) - \left( \frac{1}{\sqrt{15} - \sqrt{14}} \right) + \left( \frac{1}{\sqrt{14} - \sqrt{13}} \right) - \left( \frac{1}{\sqrt{13} - \sqrt{12}} \right) + \left( \frac{1}{\sqrt{12} - \sqrt{11}} \right)$$

Question

Evaluate the value of the expression:

$$\left( \frac{1}{\sqrt{16} - \sqrt{15}} \right) - \left( \frac{1}{\sqrt{15} - \sqrt{14}} \right) + \left( \frac{1}{\sqrt{14} - \sqrt{13}} \right) - \left( \frac{1}{\sqrt{13} - \sqrt{12}} \right) + \left( \frac{1}{\sqrt{12} - \sqrt{11}} \right)$$

Accepted Answer

Solution:
Given expression:
$$ \frac{1}{\sqrt{16}-\sqrt{15}} - \frac{1}{\sqrt{15}-\sqrt{14}} + \frac{1}{\sqrt{14}-\sqrt{13}} - \frac{1}{\sqrt{13}-\sqrt{12}} + \frac{1}{\sqrt{12}-\sqrt{11}} $$
Use the identity:
$$ \frac{1}{\sqrt{a}-\sqrt{b}} = \frac{\sqrt{a}+\sqrt{b}}{a-b} $$
Here, each denominator has consecutive numbers, so:
$$ a-b=1 $$
Therefore:
$$ \frac{1}{\sqrt{a}-\sqrt{b}}=\sqrt{a}+\sqrt{b} $$
Now simplify each term:
$$ \frac{1}{\sqrt{16}-\sqrt{15}} = 4+\sqrt{15} $$
$$ \frac{1}{\sqrt{15}-\sqrt{14}} = \sqrt{15}+\sqrt{14} $$
$$ \frac{1}{\sqrt{14}-\sqrt{13}} = \sqrt{14}+\sqrt{13} $$
$$ \frac{1}{\sqrt{13}-\sqrt{12}} = \sqrt{13}+\sqrt{12} $$
$$ \frac{1}{\sqrt{12}-\sqrt{11}} = \sqrt{12}+\sqrt{11} $$
So the expression becomes:
$$ (4+\sqrt{15})-(\sqrt{15}+\sqrt{14})+(\sqrt{14}+\sqrt{13})-(\sqrt{13}+\sqrt{12})+(\sqrt{12}+\sqrt{11}) $$
Now cancel the common terms:
$$ +\sqrt{15}-\sqrt{15}=0 $$
$$ -\sqrt{14}+\sqrt{14}=0 $$
$$ +\sqrt{13}-\sqrt{13}=0 $$
$$ -\sqrt{12}+\sqrt{12}=0 $$
Remaining value:
$$ 4+\sqrt{11} $$
Therefore,
$$ \boxed{4+\sqrt{11}} $$
Final Answer: **(a) $4+\sqrt{11}$**

Answer

Answer

Answer