2. Given $x, y \in \mathbf{R}_{+}, x^{n}+y^{n}=1(n>1, n$ $\in \mathbf{N}_{+}$). Then the maximum value of $x+4 y$ is
2. $\left(1+4 i^{n-1}\right)^{\frac{n-1}{n}}$.
$$
\begin{array}{l}
\text { Since } x^{n}+\underbrace{a^{n}+a^{n}+\cdots+a^{n}}_{n-1 \uparrow} \geqslant n a^{n-1} x, \\
y^{n}+\underbrace{b^{n}+b^{n}+\cdots+b^{n}}_{n-1 \uparrow} \geqslant n b^{n-1} y,
\end{array}
$$

where $a, b \in \mathbf{R}_{+}$, thus
$$
\begin{array}{l}
x^{n}+y^{n}+(n-1)\left(a^{n}+b^{n}\right) \geqslant n\left(a^{n-1} x+b^{n-1} y\right) . \\
\text { Let } b^{n-1}=4 a^{n-1} . \\
\text { Then } x+4 y \leqslant \frac{x^{n}+y^{n}+(n-1)\left(a^{n}+b^{n}\right)}{n a^{n-1}} .
\end{array}
$$

Thus, $x+4 y \leqslant \frac{x^{n}+y^{n}+(n-1)\left(a^{n}+b^{n}\right)}{n a^{n-1}}$.
The equality holds if and only if $x=a, y=b$.
From $x^{n}+y^{n}=1$, we get $a^{n}+b^{n}=1$.
Solving from equation (1), we get $a=\left(1+4^{\frac{n}{n-1}}\right)^{-\frac{1}{n}}$.
Therefore, from equation (2), we have $x+4 y \leqslant\left(1+4^{\frac{n}{n-1}}\right)^{\frac{n-1}{n}}$.