Square Roots in a Function!

$f(n) = \dfrac{(2n+1)+(2n-1) + \sqrt{(2n+1)(2n-1)}}{ \sqrt{2n+1} + \sqrt{2n-1}}$

$\left(\small{\text{Put }p=\sqrt{2n+1},q=\sqrt{2n-1}}\right)$

$=\dfrac{p^2+q^2+pq}{p+q}\times\dfrac{p-q}{p-q}=\dfrac{p^3-q^3}{p^2-q^2}=\dfrac{\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}}{2}$

Hence sum is:

$\sum_{n=13}^{112}\left[ \dfrac{\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}}{2}\right]$

$\small{\color{#20A900}{\text{(A Telescopic Series)}}}$

$=\dfrac{\sqrt{(2(112)+1)^3}-\sqrt{(2(13)-1)^3}}2=\dfrac{15^3-5^3}{2}=\boxed{1625}$

First, let's rationalize the function.

$f(n) = \frac{(4n + \sqrt{4n²-1})(\sqrt{2n+1} - \sqrt{2n-1})}{(\sqrt{2n+1} + \sqrt{2n-1})(\sqrt{2n+1} - \sqrt{2n-1})}$

$f(n) = \frac{(4n + \sqrt{4n²-1})(\sqrt{2n+1} - \sqrt{2n-1})}{2}$

Let $a= 2n +1$ and $b= 2n-1$ , we know $a+b = 4b$ and $ab = 4n² -1$ .

$f(n) = \frac{\left( (a+b) + \sqrt{ab} \right)(\sqrt{a} - \sqrt{b})}{2}$

$= \frac{a\sqrt{a} + \cancel{b\sqrt{a}} - \cancel{a\sqrt{b}} - b\sqrt{b} + \cancel{a\sqrt{b}} - \cancel{b\sqrt{a}}}{2}$

$= \frac{a\sqrt{a} - b\sqrt{b}}{2}$

$f(n) = \frac{1}{2} \left[ (2n+1)\sqrt{2n+1} - (2n-1)\sqrt{2n-1} \right]$ .

Now,

$f(13) = \frac{1}{2} \left[ \cancel{27\sqrt{27}} - 25\sqrt{25} \right]$

$f(14) = \frac{1}{2} \left[ \cancel{29\sqrt{29}} - \cancel{27\sqrt{27}} \right]$

$f(15) = \frac{1}{2} \left[ \cancel{31\sqrt{31}} - \cancel{29\sqrt{29}} \right]$

$\cdot \cdot \cdot \cdot \cdot \cdot$

$f(112) = \frac{1}{2} \left[ 225\sqrt{225} - \cancel{223\sqrt{223}} \right]$ .

Simplify, we have

$f(13) + f(14) + f(15) + ... + f(112) = \frac{1}{2} \left[ 225\sqrt{225} - 25\sqrt{25}\right]$

$f(13) + f(14) + f(15) + ... + f(112) = \frac{1}{2}\left(5³\right)\left(27-1\right) = 125 \times 13 = \boxed{1625}$