They had taken P versus NP farther than I'd ever seen anyone.

他们对P对NP问题的研究领先于所有人。

This is equal to np times the sum from a is equal to 0 to b.

这等于 np 乘以从 a 等于 0 到 b 的总和。

And then we can factor out this n and this p and we'll get it's equal to np times the sum from k is equal to 1 to n of-- let's see.

然后我们可以分解出这个 n 和这个 p，我们会得到它等于 np 乘以 k 的总和等于 1 到 n 的——让我们看看。

So this whole sum will then turn into np times the sum from-- OK, when k is equal to 1, that's the same thing-- when k is equal to 1, what is a equal to?

所以这个总和将变成 np 乘以总和 - 好的，当 k 等于 1 时， 那是一样的 - 当 k 等于 1 时，a 等于什么？