De Moivre–Laplace theorem By wiki

Question

https://en.m.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem. Above is the wiki link related to De Moivre–Laplace theorem. And given by the condition of proof 1,my question is why p(n, k+1)-p(n,k) is equal to np-k-q. I don’t understand the result because p(n,k+1) is binomially distribution. Hopefully there are people  can show a more detail proving step.

Answer 1

Okay,

p(n, k+1)-p(n,k)

= (_n^k+1)p^k+1q^n-k-1-(_n^k)p^kq^n-k

=n!/(k!(n-k-1)!) p^k+1q^n-k-1- n!/(k!(n-k)!)(_n^k)p^kq^n-k

=n!/(k!(n-k-1)!) p^kq^n-k-1(p/(k+1)-q/(n-k))

=n!/((k+1)!(n-k)!) p^kq^n-k-1(np-k-q)