Three. (50 points) Define the function $f(x)$ on the set of positive integers, satisfying $f(1)=\frac{4}{3}$, and $2 f(k)=f(k-1)-k - 5^{k}+2^{k} (k \geqslant 2, k \in \mathbf{N})$, try to find the analytical expression of the function $f(x)$.
Three, Solution: $\because 2 f(k)=f(k-1)-k \cdot 5^{k}+2^{k}$, $\therefore 2\left[f(k)-\frac{2^{k+1}}{3}\right]=\left[f(k-1)-\frac{2^{k}}{3}\right]-k \cdot 5^{k}$, dividing both sides by $5^{k}$, we get $2 \cdot \frac{f(k)-\frac{2^{k+1}}{3}}{5^{k}}=\frac{1}{5} \cdot \frac{f(k-1)-\frac{2^{k}}{3}}{5^{k-1}}-k$, let $g(k)=\frac{f(k)-\frac{2^{k+1}}{3}}{5^{k}}$

Then $2 g(k)=\frac{1}{5} g(k-1)-k$, adding $\frac{10 k}{9}$ to both sides, we have $2\left[g(k)+\frac{5 k}{9}\right]=\frac{1}{5}\left[g(k-1)+\frac{5(k-1)}{9}\right]+\frac{1}{9}$
Subtracting $\frac{10}{81}$ from both sides, we get $2\left[g(k)+\frac{5 k}{9}-\frac{5}{81}\right]=$ $\frac{1}{5}\left(g(k-1)+\frac{5(k-1)}{9}-\frac{5}{81}\right]$,
Let $\varphi(k)=g(k)+\frac{5 k}{9}-\frac{5}{81} \quad$ (2), then $2 \varphi(k)=$ $\frac{1}{5} \varphi(k-1)$.
$\because k$ is a natural number, $\therefore\{\varphi(k)\}$ is a geometric sequence with a common ratio of $\frac{1}{10}$, and $f(1)=\frac{4}{3}$,

From (1) and (2), we get $\varphi(1)=\frac{40}{81}, \therefore \varphi(k)=\frac{40}{81} \times$ $\left(\frac{1}{10}\right)^{k-1}$,
Thus, $g(k)=\frac{40 \times 10^{1-k}}{81}-\frac{5 k}{9}+\frac{5}{81}$,
From (1), we get $f(k)=\frac{5^{k+1}}{81}\left(8 \times 10^{1-k}-9 k+1\right)+\frac{2^{k+1}}{3}$.