So, in a problem in elementary number theory I have already shown that $$\sum_{i=1}^n {n\choose k}=2^{n} $$ and $$\sum_{i=1}^n (-1)^{k}\cdot {n\choose k}=0.$$ Using these two facts I am supposed to find the sum of $${n\choose 0} +{n\choose 2} +{n\choose 4} + \dots $$ and $${n\choose 1} +{n\choose 3} +{n\choose 5} + \dots .$$ Here's my attempt. I wish to know if this approach is legitimate or not. I have my doubts.
I figured, since $$\sum_{i=1}^n {n\choose k}=2^{n}={n\choose 0} +{n\choose 1} +{n\choose 2} + {n\choose 3} + \dots {n\choose n}$$ and $$\sum_{i=1}^n (-1)^{k}\cdot {n\choose k}=0={n\choose 0} -{n\choose 1} +{n\choose 2} - {n\choose 3} + \dots + (-1)^{n}\cdot {n\choose n}$$ then if we add these two sums we get $$\sum_{i=1}^n {n\choose k} + \sum_{i=1}^n (-1)^{k}\cdot {n\choose k}= \sum_{i=1}^n {n\choose k}\Bigl[1+(-1^{k}) \Bigr]=2^{n}=2\cdot \Biggl( {n\choose 0} +{n\choose 2} +{n\choose 4} + \dots \Biggr).$$ Thus $$\boxed{\dfrac{1}{2}\cdot \sum_{i=1}^n {n\choose k}\Bigl[1+(-1^{k}) \Bigr]={n\choose 0} +{n\choose 2} +{n\choose 4} + \dots=2^{n-1}}.$$
Similarly, $$\sum_{i=1}^n {n\choose k} - \sum_{i=1}^n (-1)^{k}\cdot {n\choose k}$$ yield $$\boxed{\dfrac{1}{2}\cdot \sum_{i=1}^n {n\choose k}\Bigl[1-(-1^{k}) \Bigr]={n\choose 1} +{n\choose 3} +{n\choose 5} + \dots=2^{n-1}}.$$
So, have I solved this problem correctly or am I missing something?
Thanks.