Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.5 The Binomial Theorem - 11.5 Assess Your Understanding - Page 856: 2

$\text{1, n}$

According to the binomial theorem, we have: $\displaystyle{n\choose k}=\dfrac{n!}{(n-k)! \ k!}$. Therefore, $\displaystyle{n\choose{0}}=\dfrac{n!}{(n-0)! \ 0 !}=\dfrac{n!}{(1) (n!)}=1$ and $\displaystyle{n\choose{1}}=\dfrac{n!}{1!(n-1)!} =\dfrac{n!}{(n-1)!}=\dfrac{n(n-1)(n-2)..(3)(2)(1)}{(n-1)(n-2)..(3)(2)(1)}=n$ Therefore, the missing terms in the given statement are: $\text{1, n}$.