[จูกัดเหลียง]

Let $\displaystyle a=x+\frac{1}{x}\rightarrow a^2+a=1$
so $\displaystyle x^2+\frac{1}{x^2}=a^2-2=-(a+1)$ and $\displaystyle x^4+\frac{1}{x^4}=a=x^6+\frac{1}{x^6}$
Thus $\displaystyle a^2=\Big(x^4+\frac{1}{x^4}\Big)\Big(x^6+\frac{1}{x^6}\Big)=x^{10}+\frac{1}{x^{10}}+x^2+\frac{1}{x^2}$
Therefore $\displaystyle x^{10}+\frac{1}{x^{10}}=2\rightarrow \Big(x^{20}+\frac{1}{x^{20}}+1\Big)\Big(x^{20}+\frac{1}{x^{20}}\Big)=6$

3.Let $$\displaystyle f(x)+f(x+1)+f(x+2)+...+f(x+2013)=\Big(\prod_{k=0}^{2013}(x+k)\Big)\Big(\sum_{k=0}^{2013} \frac{1}{(x+k)}\Big)$$
and $n\in\mathbb{N}$ such $\displaystyle \int_{1}^{2015} f(x)dx=n\cdot n!$ then find $\displaystyle \sqrt{\frac{n^4+n^2+1}{n^2+n+1}+2013}$