问题描述
执行以下归纳证明：
设d1，d2，...，dn，n至少为2，为正整数。使用数学归纳来解释为什么，如果d1 + d2 + ... + dn = 2n-2，那么必须有一个具有n个顶点的树，其度数正好是d1，d2，...，dn。 （请注意阅读此声明。与顶点度d1，d2，...，dn的任何树必须满足d1 + d2 + ... + dn = 2n-2的情况不同，尽管这也是正确的。相反，它说如果你从数字d1，d2，...，dn开始，那么你可以找到这样的树。）
Do the following inductive proof:
Let d1, d2, ..., dn, with n at least 2, be positive integers. Use mathematical induction to explain why, if d1+ d2+…+dn = 2n-2, then there must be a tree with n vertices whose degrees are exactly d1, d2, ..., dn. (Be careful with reading this statement. It is not the same as saying that any tree with vertex degrees d1, d2, ..., dn must satisfy d1+ d2+...+dn = 2n-2, although this is also true. Rather, it says that if you begin with the numbers d1, d2, ..., dn, then you can find such a tree.)
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