Question

Va rog rezolvarea!!!!! Multumesc!

Answer 1

$a)~~f(x) = x^2-x+1 \\ \\ \dfrac{x^3-1}{x^3+1} = \dfrac{(x-1)(x^2+x+1)}{(x+1)(x^2-x+1)}= \dfrac{(x-1)}{(x+1)}\cdot \dfrac{f(x+1)}{f(x)} \\ \\ f(x+1) = (x+1)^2-(x+1)+1 = x^2+2x+1-x-1+1 = x^2+x+1 \\ \\b)~~a_n = \dfrac{2^3-1}{2^3+1}\cdot \dfrac{3^3-1}{3^3+1}\cdot...\cdot \dfrac{n^3-1}{n^3+1} \\ \\ a_n = \prod\limits_{k=2}^{n} \dfrac{k^3-1}{k^3+1}=\prod\limits_{k=2}^{n} \left(\dfrac{k-1}{k+1}\cdot \dfrac{k^2+k+1}{k^2-k+1}\right) =\\ \\= \prod\limits_{k=2}^{n} \left(\dfrac{k-1}{k+1}\cdot \dfrac{(k+1)^2-(k+1)+1}{k^2-k+1}\right) =$

$= \dfrac{\prod\limits_{k=2}^{n}(k-1)}{\prod\limits_{k=2}^{n}(k+1)}\cdot \dfrac{\prod\limits_{k=2}^{n}[(k+1)^2-(k+1)+1]}{\prod\limits_{k=2}^{n}(k^2-k+1)} =$

$=\dfrac{\prod\limits_{k=1}^{n-1}k}{\prod\limits_{k=3}^{n+1}k} \cdot \dfrac{\prod\limits_{k=3}^{n+1}(k^2-k+1)}{\prod\limits_{k=2}^{n}(k^2-k+1)} =\\ \\ = \dfrac{1\cdot 2\cdot \prod\limits_{k=3}^{n-1}k}{\prod\limits_{k=3}^{n-1}(k)\cdot n(n+1)}\cdot \dfrac{\prod\limits_{k=3}^{n}(k^2-k+1)\cdot \Big[(n+1)^2-(n+1)+1\Big]}{(2^2-2+1)\cdot \prod\limits_{k=3}^{n}(k^2-k+1)} = \\ \\ = \dfrac{2}{n(n+1)}\cdot \dfrac{(n+1)^2-(n+1)+1}{4-2+1} = \dfrac{2\cdot (n^2+2n+1-n-1+1)}{3n(n+1)}= \\ \\ = \boxed{\dfrac{2(n^2+n+1)}{3n(n+1)}}$