Assuming $0\leq \lambda \leq \frac{1}{2}$ show that: $$1 = (1 + (1- \lambda))^n = \sum_{k=0}^n \binom{n}{k} \lambda^k (1- \lambda)^{n-k}.$$
I understand the general procedure of the binomial theorem, namely that: $$ (1 + (1- \lambda))^n = \binom{n}{0}1^n (1- \lambda)^0 + \binom{n}{1}1^{n-1}(1- \lambda)^1 + \dots + \binom{n}{n-1}1^1 (1- \lambda)^{n-1} + \binom{n}{n}1^0 (1- \lambda)^n= \sum_{k=0}^n {n \choose k} 1^{n - k} (1- \lambda)^k.$$
What gives me trouble on the one hand is the "trick" to get to $$1 = (1 + (1- \lambda))^n,$$
on the other I do not know how to switch arround the terms to get $$\sum_{k=0}^n \binom{n}{k} \lambda^k (1- \lambda)^{n-k}$$ instead of $$\sum_{k=0}^n {n \choose k} 1^k (1- \lambda)^{n - k}$$.
I am an EE undergrad just getting started with mathematics. Thanks.
No, the basic equation is $$1 = (\lambda + (1-\lambda))^n.$$ The right-hand side can be expanded as you've shown.