So we have
$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k} = (0.3+0.7)^n=1^n=1$
Now I want to solve
$\sum _{k=\frac{n}{2}+2}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$
This is what I tried doing
$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{n}{2}+2}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+2-k} +\sum _{k=0}^{\frac{n}{2}+1}\binom{\frac{n}{2}+1\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+1-k}=1 $
$\sum _{k=\frac{n}{2}+2}^n\binom{\frac{n}{2}+2\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+2-k}= 1-\sum _{k=0}^j\binom{j\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{j-k}, j=\frac{n}{2}+1$
$\sum _{k=\frac{n}{2}+2}^n\binom{\frac{n}{2}+2\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+2-k}= 1-1^{\frac{n}{2}+1}=0$
Now this is obviously wrong so I am not sure how to make it right.
Hint: $$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=1$$ if $n$ is even $$S=\sum _{k=0}^{\frac{n}{2}-1}\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{n}{2}+1}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$$
leaving one term in between $$\binom{n\:}{\frac{n}{2}}\:\left(0.3\right)^k\left(0.7\right)^{n-\frac{n}{2}}$$
$$S+S+\binom{n\:}{\frac{n}{2}}\:\left(0.3\right)^k\left(0.7\right)^{n-\frac{n}{2}}=1$$ if $n$ is odd
$$\sum _{k=0}^{\frac{2n-1}{2}}\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{2n+1}{2}}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$$