48. A person's average speed going up the mountain (from the foot to the top) is $V_{1}$, and the average speed going down the mountain (from the top to the foot, returning the same way) is $V_{2}$,

From the start of the climb to the top and immediately descending back to the foot, the average speed for the entire journey is $V$, given that $V_{2}=k V_{1}$, $V=m V_{1}$, $V=n V_{2}$, then, $m=$ $\qquad$ (expressed as an algebraic expression containing only the letter $k$); $m+n=$ $\qquad$
Answer: $\frac{2 k}{1+k} ; 2$
Solution: Let the distance from the foot of the mountain to the top be $s$, then the time taken to go up the mountain is $\frac{s}{V_{1}}$, and the time taken to go down the mountain is $\frac{s}{V_{2}}$, so
$$
V=2 s,\left(\frac{s}{V_{1}}+\frac{s}{V_{2}}\right)=\frac{2 V_{1} V_{2}}{V_{1}+V_{2}},
$$

Since $V_{2}=k V_{1}$, we have $V=\frac{2 V_{1}\left(k V_{1}\right)}{V_{1}+k V_{1}}=\frac{2 k V_{1}}{1+k}$, thus $m=\frac{V}{V_{1}}=\frac{2 k}{1+k}$.
From
$$
V=n V_{2} \text {, }
$$

we get
$$
n=\frac{V}{V_{2}}=\frac{V}{k V_{1}}=\frac{1}{k}, \frac{2 k}{1+k}=\frac{2}{1+k},
$$

then
$$
m+n=\frac{2 k}{1+k}+\frac{2}{1+k}=\frac{2 k+2}{1+k}=2
$$