Prove that two qualtities are equal (possibly using binomial expansion)

Question

How do you prove that $$\sum_{i=0}^{\frac{n}{2}} {n \choose 2i}p^{2i}(1-p)^{n-2i} = \frac{1+(1-2p)^n}{2}$$

I tried using binomial expansion but I can't because the upper limit is isn't n.

note: n cannot be odd

Answer 1

Denote the sum as $S_n$ as $$S_n=\sum_{i=0}^{n/2} {n \choose 2i} p^{2i}(1-p)^{n-2i}=(1-p)^n \sum_{k=0}^{n/2} {n \choose 2i} \left(\frac{p}{1-p}\right)^{2i}~~~~~~(1)$$ Note yhe identity $$\frac{(1+x)^n+(1-x)^n}{2} = \sum_{k=0}^{n/2} {n \choose 2k} x^{2k}~~~~(2).$$ Then (1) becomes $$ S_n=(1-p)^n \frac{[1+p/(1-p)]^n+[1-p/(1-p)]^n}{2}=\frac{1+(1-2p)^n}{2}.$$

Answer 2

Consider $\left[p+\left(1-p\right)\right]^n$ and $\left[p-\left(1-p\right)\right]^n$: $$\left[p+\left(1-p\right)\right]^n=\sum_{k=0}^n \binom{n}{k}p^k \left(1-p\right)^{n-k}\\\left[p-\left(1-p\right)\right]^n=\sum_{k=0}^n \binom{n}{k}p^k \left(1-p\right)^{n-k}\left(-1\right)^{n-k}=\sum_{k=0}^n \binom{n}{k}p^k \left(1-p\right)^{n-k}\left(-1\right)^{k}$$ Find the average of this two expression: $$L.H.S.=\dfrac{\left[p+\left(1-p\right)\right]^n+\left[p-\left(1-p\right)\right]^n}{2}=\dfrac{1+\left(1-2p\right)^n}{2}\\R.H.S.=\sum_{k=0}^n \binom{n}{k}p^k \left(1-p\right)^{n-k}\left[\dfrac{1+\left(-1\right)^{k}}{2}\right]\\ \because\dfrac{1+\left(-1\right)^{k}}{2}=0\text{ when }k\text{ is odd}\\\therefore R.H.S.=\sum_{i=0}^\frac{n}{2} \binom{n}{2i}p^{2i}\left(1-p\right)^{n-2i}\\ \boxed{\sum_{i=0}^\frac{n}{2} \binom{n}{2i}p^{2i}\left(1-p\right)^{n-2i}=\dfrac{1+\left(1-2p\right)^n}{2}}$$