100 followers problem

Let $\large S = \displaystyle\sum_{x=4}^{11} \dfrac{x^9 + 7x^8 + x^7 -69x^6 -64x^5 + 324x^4 +464x^3 -560x^2 -1212x -320}{x^{10}+4x^9 -12x^8 -64x^7 + 32x^6 + 384x^5 + 128x^4 -1024x^3 - 768x^2 + 1024x +1024}$

$$

This can be written as

$\large \displaystyle\sum_{x=4}^{11} \dfrac{x^9 + 6x^8 -64 x^6 - 96x^5 + 192x^4 + 512x^3 - 768x - 512 \\ + ( x^8 -16x^6 + 96x^4 -256x^2 + 256) \\ + ( x^7 + 10x^6 + 36x^5 + 40x^4 - 80x^3 - 288x^2- 380x - 128) \\ + ( x^6 - 4x^5 - 4x^4 + 32x^3 -64x + 64 -16x^2) }{x^{10}+4x^9 -12x^8 -64x^7 + 32x^6 + 384x^5 + 128x^4 -1024x^3 - 768x^2 + 1024x +1024}$

$\\$

After factorizing:

$\large S = \displaystyle\sum_{x=4}^{11} \dfrac{( x+2)^4 ( x-2)^3 (x+2)^2 \\ + (( x+2)^4 ( x-2)^3 (x-2)) \\ +( x+2)^4 (x+2)^2 (x-2) \\ +( ( x-2)^3 (x+2)^2 (x-2))}{( x+2)^4 ( x-2)^3 (x+2)^2 (x-2)}$

$\\$

This gives us .

$\large S = \displaystyle\sum_{x=4}^{11} \dfrac{1}{(x+2)^4} + \dfrac{1}{(x-2)^3} + \dfrac{1}{(x+2)^2} + \dfrac{1}{(x-2)}$

$\\$

On simplifying, we get

$\dfrac{1}{6^4} + \dfrac{1}{2^3} + \dfrac{1}{6^2} + \dfrac{1}{2} \\\cdots \\\cdots \\\cdots \\\cdots$

Therefore $S$ is telescopic and on simplification gives the answer $\large \boxed{ \color{#E81990}{\boxed{\color{#3D99F6}{2.1346}}}}$