Show that for any number n and real number x that

\sum_{i=0}^n l_i(x) = 1.

My attempt at a proof is that we should argue by induction, taking n=1 as a base case, which is easy to show:

\frac{x-x_1}{x_0-x_1}+\frac{x-x_0}{x_1-x_0}

=\frac{x_1-x+x-x_0}{x_1-x_0}

=\frac{x_1-x_0}{x_1-x_0}=1

So now let's assume that \displaystyle\sum_{i=0}^n l_i(x)=1. Then:

\displaystyle\sum_{i=0}^{n+1}l_i(x)
=\displaystyle\sum_{i=0}^n l_i(x) + l_{n+1}(x)
=(x-x_{n+1})\frac{1}{????} + l_{n+1}(x)

And it's finding out what that fraction is going to be over (as I know that sum is not just 1 due to the (x_i-x_{n+1}) terms that will be in the bottom. I think once I figure that one out the proof should fall out nicely (unless of course I'm making it way too hard and it's easier to prove directly).