略．

本题可证明 $F_n(\dfrac 12)F_n(1)<0$，然后证明 $F_n(x)$ 在 $(\dfrac 12,1)$ 上单调，即可得到题中结论，然后令 $F_n(x_n)=0$，即可得到 $x_n=\dfrac 12+\dfrac 12x_n^{n+1}$．由题意，\[F_n\left(x\right)=f_n\left(x\right)-2=1+x+x^2+\cdots+x^n-2,\]则 $F_n\left(1\right)=n-1>0$．\[\begin{split}F_n\left(\dfrac 12\right)&=1+\dfrac 12+\left(\dfrac 12\right)^2+\cdots+\left(\dfrac 12\right)^n-2\\&\overset{\left[a\right]}=\dfrac {1-\left(\dfrac 12\right)^{n+1}}{1-\dfrac 12}-2\\&=-\dfrac {1}{2^n}<0,\end{split}\]（推导中用到：[a]）
所以 $F_n\left(x\right)$ 在 $\left(\dfrac 12,1\right)$ 内至少存在一个零点．
又 $F'_n\left(x\right)=1+2x+\cdots+nx^{n-1}>0$，故 $F_n\left(x\right)$ 在 $\left(\dfrac 12,1\right)$ 内单调递增，
所以 $F_n\left(x\right)$ 在 $\left(\dfrac 12,1\right)$ 内有且仅有一个零点 $x_n$．
因为 $x_n$ 是 $F_n\left(x\right)$ 的零点，所以 $F_n\left(x_n\right)=0$，即\[\dfrac {1-x_n^{n+1}}{1-x_n}-2=0,\]故 $x_n=\dfrac 12+\dfrac 12x_n^{n+1}$．

当 $x=1$ 时，$f_n\left(x\right)=g_n\left(x\right)$；当 $x\neq 1$ 时，$f_n\left(x\right)<g_n\left(x\right)$．

可构造函数 $h(x)=f_n(x)-g_n(x)$，然后通过导数研究它的单调性，继而得到 $h(x)$ 和 $0$ 的大小关系，从而得出 $f_n(x)$ 和 $g_n(x)$ 的大小关系．方法一：
由题设，$g_n\left(x\right)=\dfrac {\left(n+1\right)\left(1+x^n\right)}{2}$．
可用作差法比较 $f_n\left(x\right)$ 和 $g_n\left(x\right)$ 的大小．设\[\begin{split}h\left(x\right)&=f_n\left(x\right)-g_n\left(x\right)\\&=1+x+x^2+\cdots+x^n-\dfrac {\left(n+1\right)\left(1+x^n\right)}{2} , x>0 .\end{split}\]当 $x=1$ 时，$f_n\left(x\right)=g_n\left(x\right)$．
当 $x\neq 1$ 时，\[h'\left(x\right)=1+2x+\cdots+nx^{n-1}-\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}.\]当 $x> 1$ 时，\[\begin{split}h'\left(x\right)&<x^{n-1}+2x^{n-1}+\cdots+nx^{n-1}-\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}\\&=\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}-\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}=0.\end{split}\]当 $0<x< 1$ 时，\[\begin{split}h'\left(x\right)&>x^{n-1}+2x^{n-1}+\cdots+nx^{n-1}-\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}\\&=\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}-\dfrac {n\left(n+1\right)}{2}\cdot x^{n-1}=0.\end{split}\]所以 $h\left(x\right)$ 在 $\left(0,1\right)$ 上递增，在 $\left(1,+\infty\right)$ 上递减．
所以 $h\left(x\right)<h\left(1\right)=0$，即 $f_n\left(x\right)<g_n\left(x\right)$．
综上所述，当 $x=1$ 时，$f_n\left(x\right)=g_n\left(x\right)$；当 $x\neq 1$ 时，$f_n\left(x\right)<g_n\left(x\right)$．
方法二：
由题设，$f_n\left(x\right)=1+x+x^2+\cdots+x^n$，$g_n\left(x\right)=\dfrac {\left(n+1\right)\left(1+x^n\right)}{2},x>0$．
当 $x=1$ 时，$f_n\left(x\right)=g_n\left(x\right)$．
当 $x\neq 1$ 时，用数学归纳法可以证明 $f_n\left(x\right)<g_n\left(x\right)$．
① 当 $n=2$ 时，$f_2\left(x\right)-g_2\left(x\right)=-\dfrac 12\left(1-x\right)^2<0$，
所以 $f_2\left(x\right)<g_2\left(x\right)$ 成立．
② 假设 $n=k\left(k\geqslant 2\right)$ 时，不等式成立，即 $f_k\left(x\right)<g_k\left(x\right)$．
那么，当 $n=k+1$ 时，\[\begin{split}f_{k+1}\left(x\right)&=f_k\left(x\right)+x^{k+1}\\&<g_k\left(x\right)+x^{k+1}\\&=\dfrac {\left(k+1\right)\left(1+x^k\right)}{2}+x^{k+1}\\&=\dfrac {2x^{k+1}+\left(k+1\right)x^{k}+k+1}{2}.\end{split}\]又\[\begin{split}g_{k+1}\left(x\right)-\dfrac {2x^{k+1}+\left(k+1\right)x^{k}+k+1}{2}=\dfrac {kx^{k+1}-\left(k+1\right)x^k+1}{2}.\end{split}\]令 $h_k\left(x\right)=kx^{k+1}-\left(k+1\right)x^k+1\left(x>0\right)$，则\[\begin{split}h'_k\left(x\right)&=k\left(k+1\right)x^{k}-k\left(k+1\right)x^{k-1}\\&=k\left(k+1\right)x^{k-1}\left(x-1\right).\end{split}\]所以当 $0<x<1$ 时，$h'_k\left(x\right)<0$，$h_k\left(x\right)$ 在 $\left(0,1\right)$ 上递减；
当 $x>1$ 时，$h'_k\left(x\right)>0$，$h_k\left(x\right)$ 在 $\left(1,+\infty\right)$ 上递增．
所以 $h_k\left(x\right)>h_k\left(1\right)=0$，
从而 $g_{k+1}\left(x\right)>\dfrac {2x^{k+1}+\left(k+1\right)x^{k}+k+1}{2}$．
故 $f_{k+1}\left(x\right)<g_{k+1}\left(x\right)$，即 $n=k+1$ 时不等式也成立．
由 $ ① $ 和 $ ② $ 知，对一切 $n\geqslant 2$ 的整数，都有 $f_n\left(x\right)<g_n\left(x\right)$．