n> 2 ，调用 getSum（n）以递归方式调用自身两次。每个调用都可以进一步递归。方法调用的总数按 2 ^ n 进行缩放， 2 ^ 50 是一个非常大的数字。这种不良的缩放反映了这样一个事实：简单的递归方法最终会不必要地重新计算相同的结果（例如fib（4））很多次，这就是为什么你的程序在增加 n 。

For n > 2, an invocation of your getSum(n) recursively invokes itself twice. Each of those invocations may recurse further. The total number of method invocations scales as 2^n, and 2^50 is a very large number. This poor scaling reflects the fact that the simple-minded recursive approach ends up needlessly recomputing the same results (e.g. fib(4)) a great many times, and it is why your program slows down so rapidly as you increase n.

超过数据类型 int的限制后，您获得的负返回值。您可以使用更宽的数据类型获得更大的限制，大概是 long 。如果这还不够，那么你需要去一些类似 BigInteger 的东西，而且会有很大的性能损失。

The negative return value you get after a certain point arises from exceeding the limits of data type int. You could get a larger limit with a wider data type, presumably long. If that's not enough then you would need to go to something like BigInteger, at a substantial performance penalty.