Function Challenge 2!

$f\left( x \right) =\frac { { x }^{ 2 }+12x+18 }{ { x }^{ 2 }-8x-2 } \\ By\quad observing\quad the\quad above\quad equation\quad and\quad the\quad extra\quad term\quad \\ invoving\quad 'n'\quad in\quad the\quad denominator\quad of\quad the\quad question,\quad \\ one\quad can\quad conclude\quad that\quad the\quad above\quad equation\quad must\quad \\ be\quad modified\quad somehow\quad so\quad as\quad to\quad cancel\quad out\quad some\quad terms.\\ \\ f\left( x \right) =\frac { { x }^{ 2 }+12x+18+(18-18) }{ { x }^{ 2 }-8x+(16-16)-2 } \\ f\left( x \right) =\frac { { (x }+6)^{ 2 }-18 }{ { (x-4) }^{ 2 }-18 } \\ Now\quad that\quad looks\quad a\quad lot\quad more\quad like\quad the\quad denominator\quad of\quad the\quad \\ second\quad term\quad under\quad the\quad root\quad in\quad the\quad denominator.\\ Using\quad the\quad above\quad equation,\quad one\quad can\quad see\quad that\quad the\quad \\ denominator\quad of\quad f\left( 11 \right) \quad cancels\quad out\quad the\quad numerator\quad of\quad f\left( 1 \right) .\\ And\quad continues\quad until\quad the\quad denominator\quad of\quad f\left( 100 \right) \quad cancels\quad out\quad \\ the\quad numerator\quad of\quad f\left( 90 \right) .\\ Therefore,\quad the\quad remaining\quad terms\quad are:\\ \displaystyle\prod _{ x=1 }^{ 100 }{ \frac { { (x }+6)^{ 2 }-18 }{ { (x-4) }^{ 2 }-18 } } = \prod _{ x=1 }^{ 10 }{ [\frac { 1 }{ { (x-4) }^{ 2 }-18 } ] } \prod _{ x=91 }^{ 100 }{ { [(x }+6)^{ 2 }-18 } ]\\ Changing\quad the\quad variable\quad x\quad into\quad (x+6),\\ \displaystyle\prod _{ x=91 }^{ 100 }{ { [(x }+6)^{ 2 }-18 } ]=\prod _{ x=97 }^{ 106 }{ { [x }^{ 2 }-18 } ]\\ Which\quad cancels\quad out\quad with\quad\displaystyle \prod _{ n=97 }^{ 106 }{ \frac { 1 }{ n^{ 2 }-18 } } .\\ We\quad are\quad left\quad with:\\ \displaystyle \prod _{ x=1 }^{ 10 }{ [\frac { 1 }{ { (x-4) }^{ 2 }-18 } ] } =\frac { 1 }{ (-{ 9) }^{ 2 } } \times \frac { 1 }{ { (-14) }^{ 2 } } \times \frac { 1 }{ { (-17) }^{ 2 } } \times \frac { 1 }{ { { (-18) }^{ 2 } } } \times \frac { 1 }{ (-2) } \times \frac { 1 }{ (-7) } \\ Answer=\frac { 7 }{ 306\sqrt { 14\times \displaystyle\prod _{ x=1 }^{ 10 }{ [\frac { 1 }{ { (x-4) }^{ 2 }-18 } ] } } } =\frac { 7 }{ 306\times \frac { 1 }{ ({ 9) } } \times \frac { 1 }{ { (14) } } \times \frac { 1 }{ { (17) } } \times \frac { 1 }{ { { (18) } } } } =\boxed { 882 }$

First, we'll look at $\displaystyle \prod_{x=1}^{100}{f(x)}$ , In the function $f(x)=\frac{x^2+12x+18}{x^2-8x-2}$ , substitute $x$ with $x+10$ then you have $f(x+10)=\frac{x^2+32x+238}{x^2+12x+18}$ , notice the denominator of $f(x+10)$ and the numerator of $f(x)$ are the same, so when they are multiplied together, which is $f(x)\times f(x+10)$ , we could cancel out the numerator of $f(x)$ with the denominator of $f(x+10)$ . Similarly, we could cancel out the numerator of $f(x+1)$ with the denominator of $f(x+11)$ , and so as the numerator of $f(x+2)$ with the denominator of $f(x+12)$ and so on. Thus, the numerator of the product is left with numerators from $f(91)$ to $f(100)$ , and the denominator of the product is left with denominators from $f(1)$ to $f(10)$ .

Now let's concentrate on the denominator of the product. The denominator of the product is $\displaystyle \prod_{x=1}^{10}{(x^2-8x-2)}$ . Notice that if we substitute $x$ with $x+4$ , we get $\displaystyle \prod_{x=-3}^{6}{(x^2-18)}$ and can easily get the product is $9^2\times 14^3\times 17^2\times 18^2$ .

On the other hand, the numerator of the product is $\displaystyle \prod_{x=91}^{100}{(x^2+12x-18)}$ , we substitue $x$ with $x-6$ and we have $\displaystyle \prod_{x=97}^{106}{(x^2-18)}=\displaystyle \prod_{n=97}^{106}{(n^2-18)}$ .

Hence, $\begin{aligned} \frac{7}{306 \sqrt{\displaystyle 14 \left (\prod_{x=1}^{100}{f(x)} \right) \left(\prod_{n=97}^{106}{\frac{1}{n^2-18}}\right)}} &= \frac{7}{306 \sqrt{\frac{1}{9^2\times 14^2\times 17^2\times 18^2}}} \\ &=882 \\ \end{aligned}$