\displaystyle\sum_{\substack{
 1\le k\le n \\
 k \mathrm{ neparan}
 }}{n \choose k}(1/6)^k(5/6)^{n-k} = \displaystyle\sum_{\substack{
 1\le k\le n \\
 k \mathrm{ neparan}
 }}{n \choose k}\frac{5^{n-k}}{6^n}