$\newcommand\E{\mathbb E}$Suppose we toss a coin $n$ times, and let the r.v. $X$ denote the number of heads we see. Moreover, let $X_{i} = 1$ if the $i$-th toss is a heads, otherwise $0$.
$$\mathrm{Var}(X) = \E(X^{2}) - (\E(X))^{2} = \sum_{i=1}^{n} \Pr(X_{i} = 1)(1^{2}) - \left(\frac{n}{2}\right)^{2} = \frac{n}{2} - \frac{n^{2}}{4}$$
Thus, if $n=100$, this would mean the variance is $50 - 2500 = -2450$...
What am I missing here?
$\DeclareMathOperator\E{\mathbb E}$This "equality": $$ \E(X^2)\ ``="\sum_{i=1}^n \Pr(X_i = 1) (1^2) $$ is not correct. That's saying $$ \left( \sum_{i=1}^n a_i \right)^2\ ``=" \sum_{i=1}^n a_i^2 ,$$ when it should actually be $$ \left( \sum_{i=1}^n a_i \right)^2 = \sum_{i=1}^n a_i^2 + \sum_{i \ne j} a_i a_j .$$
The correct version is then \begin{align} \E(X^2) &= \E \left( \sum_{i=1}^n X_i^2 + \sum_{i \ne j} X_i X_j \right) \\&= \sum_{i=1}^n \E\left[ X_i^2 \right] + \sum_{i \ne j} \E\left[ X_i X_j \right] \\&= \sum_{i=1}^n \left( \Pr\left(X_i = 0 \right) 0^2 + \Pr\left( X_i = 1 \right) 1^2 \right) + \sum_{i \ne j} \E\left[ X_i \right] \E\left[ X_j \right] \\&= \sum_{i=1}^n \frac12 + \sum_{i \ne j} \frac{1}{2} \frac{1}{2} \\&= \frac{n}{2} + \frac{1}{4} n (n-1) \\&= \frac14 n^2 + \frac14 n ,\end{align} so that $$ \mathrm{Var}(X) = \frac14 n^2 + \frac14 n - \frac14 n^2 = \frac{n}{4} .$$
