Testo nascosto, fai click qui per vederlo
Let $2016=a$ and $2017=b$. We want the value of $\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}-\frac{8a^7}{a^8-b^8}$. Note that $\frac{4a^3}{a^4+b^4}=\frac{4a^3(a^4-b^4)}{(a^4+b^4)(a^4-b^4)}=\frac{4a^7-4a^3b^4}{a^8-b^8}$, so the last two fractions combine to $\frac{-4a^7-4a^3b^4}{a^8-b^8}=\frac{-4a^3(a^4+b^4)}{(a^4+b^4)(a^4-b^4)}=-\frac{4a^3}{a^4-b^4}$. Then we want $\frac{1}{a+b}+\frac{2a}{a^2+b^2}-\frac{4a^3}{a^4-b^4}=\frac{1}{a+b}+\frac{2a(a^2-b^2)-4a^3}{a^4-b^4}=\frac{1}{a+b}+\frac{-2a(a^2+b^2)}{(a^2-b^2)(a^2+b^2)}=\frac{1}{a+b}-\frac{2a}{a^2-b^2}=\frac{a-b-2a}{a^2-b^2}=-\frac{a+b}{(a-b)(a+b)}=-\frac{1}{a-b}=-\frac{1}{2016-2017}=1$.