Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$ Binomial problems: Mean and standard deviation
Suppose that the New England Colonials baseball team is equally likely to win any particular game as not to win it. Suppose also that we choose a random sample of Colonials games.
Estimate the number of games in the sample that the Colonials win by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.
Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.
I've calculated the mean to be $10 (np=20*0.5)$. The short cut for calculating the variance of a binomial distribution is $np(1-p)$, but can you show me how to use the standard discrete distribution formula for variance ($Var(X) = E(X^2) - \mu^2$)to calculate variance in this case?
 A: \begin{align}
\textrm{Var}\left[X\right] &= \sum_{i=0}^{n}\left(i-np\right)^2\binom{n}{i}p^i(1-p)^{n-i}\\
&= \sum_{i=0}^{n}\left(i^2-2inp+\left(np\right)^2\right)\binom{n}{i}p^i(1-p)^{n-i}\\
&= \sum_{i=0}^{n}i^2\binom{n}{i}p^i(1-p)^{n-i}-2np\sum_{i=0}^{n}i\binom{n}{i}p^i(1-p)^{n-i}+(np)^2\sum_{i=0}^{n}\binom{n}{i}p^i(1-p)^{n-i}\\
&= \sum_{i=0}^{n}i^2\binom{n}{i}p^i(1-p)^{n-i}-2(np)^2+(np)^2\\
&= \sum_{i=0}^{n}i^2\binom{n}{i}p^i(1-p)^{n-i}-(np)^2\\
&= \sum_{i=0}^{n}ni\binom{n-1}{i-1}p^i(1-p)^{n-i}-(np)^2\\
&= np\sum_{i=0}^{n}i\binom{n-1}{i-1}p^{i-1}(1-p)^{n-i}-(np)^2\\
&= np\sum_{i=1}^{n}i\binom{n-1}{i-1}p^{i-1}(1-p)^{n-i}-(np)^2\\
&= np\sum_{j=0}^{n-1}(j+1)\binom{n-1}{j}p^{j}(1-p)^{n-1-j}-(np)^2\\
&= np\left(\sum_{j=0}^{n-1}j\binom{n-1}{j}p^{j}(1-p)^{n-1-j}+\sum_{j=0}^{n-1}\binom{n-1}{j}p^{j}(1-p)^{n-1-j}\right)-(np)^2\\
&= np\left((n-1)p+1\right)-(np)^2\\
&= np\left(np-p+1\right)-(np)^2\\
&= (np)^2-np^2+np-(np)^2\\
&= np-np^2\\\
&= np(1-p)\\
\end{align}
EDIT: I misunderstood the question.
The variance can be calculated as follows:
$\textrm{Var}(X)=\left(0^2\times P(X=0)+1\times P(X=1)+\cdots+20^2\times P(X=20)\right)-10^2$
