This might be a really stupid question... but still here we go:
For the Variance we have two formulas: (i)$$ \sigma^2:=\sum_{k=0}^n (k-\mu)^2p = \sum_{k=0}^n (k^2-2npk+n^2p^2)p= (\sum_{k=0}^n k^2- \sum_{k=0}^n2npk+\sum_{k=0}^n n^2p^2)p= \sum_{k=0}^nk^2p - 2np\sum_{k=0}^nkp +n^2p^2 = \sum_{k=0}^nk^2p - 2n^2p^2+n^2p^3 $$ with $\mu=np$ being the Expected Value and also (ii) $$ Var(X)=E(X^2)-(E(X))^2 = \sum_{k=0}^nk^2p - (np)^2 $$ Now $n^2p^2+n^2p^3\neq (np)^2$ obviously, so where is my mistake?
The correct method to obtain the variance in both ways begins with either $$\text{Var}(X)=E((X-\mu)^2)=\sum_{k=0}^n\left((k-np)^2\binom{n}{k}p^k(1-p)^{n-k}\right)$$ or $$\text{Var}(X)=E(X^2)-\mu^2=\sum_{k=0}^n\left(k^2\binom{n}{k}p^k(1-p)^{n-k}\right)-(np)^2$$