Variance of Negative Binomial Distribution

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Most sources say it's $\frac{r(1-p)}{p^2}$ but I can't seem to get it. What am I doing wrong?

Let $X$ =$X_1 + X_2 + \cdots + X_r$.

$Var(X) = E(X^2) - E(X)^2$

$Var(X_1 + X_2 + \cdots + X_r) = E((X_1 + X_2 + \cdots + X_r)^2) - (E(X_1 + X_2 + \cdots + X_r))^2$

Then,

$E((X_1 + X_2 + \cdots + X_r)^2)$

$ = E(\sum_{i,j} X_iX_j)$

$ = \sum_{i, j} E(X_iX_j)$

$ = \sum_{i, j} E(X_i)E(X_j)$

since $X_i$ and $X_j$ are independent for all $i, j$.

$ = (r^2) E(X_i)E(X_j)$

and since $E(X_i) = \frac{1}{p}$ for all $i$, then our expression equals $\frac{r^2}{p^2}$

which is clearly wrong.