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Algebra Difficulty 4.8 AIME Find the answer

3. f(x)f(x) is a function defined on (0,1)(0,1), for any 1<x<y<+∞1<x<y<+\infty, we have
f(1x)−f(1y)=f(x−y1−xy). f\left(\frac{1}{x}\right)-f\left(\frac{1}{y}\right)=f\left(\frac{x-y}{1-x y}\right) .

Let an=f(1n2+5n+5)(n∈N+)a_{n}=f\left(\frac{1}{n^{2}+5 n+5}\right)\left(n \in \mathbf{N}_{+}\right). Then a1+a2+⋯+a8=()a_{1}+a_{2}+\cdots+a_{8}=(\quad).

Pick one

Solution

3. C.

Notice,
an=f(1n2+5n+5)=f((n+2)−(n+3)1−(n+2)(n+3))=f(1n+2)−f(1n+3). \begin{array}{l} a_{n}=f\left(\frac{1}{n^{2}+5 n+5}\right) \\ =f\left(\frac{(n+2)-(n+3)}{1-(n+2)(n+3)}\right) \\ =f\left(\frac{1}{n+2}\right)-f\left(\frac{1}{n+3}\right) . \end{array}

Then a1+a2+⋯+a8a_{1}+a_{2}+\cdots+a_{8}
=f(13)−f(14)+f(14)−f(15)+⋯+f(110)−f(111)=f(13)−f(111)=f(14). \begin{array}{l} =f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)+f\left(\frac{1}{4}\right)-f\left(\frac{1}{5}\right)+\cdots+ \\ f\left(\frac{1}{10}\right)-f\left(\frac{1}{11}\right) \\ =f\left(\frac{1}{3}\right)-f\left(\frac{1}{11}\right)=f\left(\frac{1}{4}\right) . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.