数学归纳法基例n=1时，如何处理求和项(2+3+...+n)？

Hey there, let's clear up this confusion with your induction base case! The key here is understanding empty sums—a straightforward math convention that's easy to miss when you're first working through these proofs.

When n=1, your summation term is $(2+3+...+n)$ which translates to $(2+3+...+1)$. Notice that the starting value of the sum (2) is actually larger than the ending value (1). In math, we call this an empty sum—there are no terms to add together! By definition, an empty sum equals 0 (since adding nothing to a total leaves it exactly as it was).

Let's plug this into your equation to verify the base case:

• Left-hand side (LHS) when n=1:
  1 + 2*(empty sum) + (1+1) = 1 + 2*0 + 2 = 3
• Right-hand side (RHS) when n=1:
  (1+1)^2 - 1 = 2² - 1 = 4 - 1 = 3

Perfect, LHS equals RHS, so your base case holds solid!

As a quick tip for the inductive step: Assume the equation is true for some integer k ≥ 1, then when you move to n=k+1, the summation will become $(2+3+...+k+(k+1))$—just the sum from your inductive hypothesis plus $(k+1)$. Use that to rewrite the left-hand side and simplify it to match the right-hand side $( (k+2)^2 - 1 )$.

内容的提问来源于stack exchange，提问作者SteveK3223