[proofplan] We expand $\operatorname{Var}(\sum X_i)$ using the computational formula, distribute the square of the sum into diagonal and cross terms, and show that each cross term $\mathbb{E}[X_i X_j] - \mathbb{E}[X_i]\mathbb{E}[X_j]$ vanishes by pairwise independence. The result is proved by induction on $n$, with the base case $n = 2$ containing the key computation. [/proofplan] [step:Reduce to the case $n = 2$ by induction] We proceed by induction on $n$. The case $n = 1$ is the identity $\operatorname{Var}(X_1) = \operatorname{Var}(X_1)$. For $n \ge 2$, write $S_n = S_{n-1} + X_n$ where $S_{n-1} = X_1 + \cdots + X_{n-1}$. If we establish the result for $n = 2$ (any two pairwise independent random variables), then the inductive step follows: by the inductive hypothesis, $\operatorname{Var}(S_{n-1}) = \sum_{i=1}^{n-1} \operatorname{Var}(X_i)$, and then \begin{align*} \operatorname{Var}(S_n) = \operatorname{Var}(S_{n-1}) + \operatorname{Var}(X_n) = \sum_{i=1}^n \operatorname{Var}(X_i), \end{align*} provided we verify that $S_{n-1}$ and $X_n$ satisfy the hypothesis of the two-variable case. Since pairwise independence of $X_1, \ldots, X_n$ implies $\operatorname{Cov}(S_{n-1}, X_n) = 0$ (as we show in the next step), this suffices. [/step] [step:Prove the $n = 2$ case: expand $\operatorname{Var}(X + Y)$ and show the cross term vanishes] Let $X$ and $Y$ be pairwise independent with finite expectations. Let $\mu_X = \mathbb{E}[X]$ and $\mu_Y = \mathbb{E}[Y]$. Then \begin{align*} \operatorname{Var}(X + Y) &= \mathbb{E}[(X + Y)^2] - (\mathbb{E}[X + Y])^2 \\ &= \mathbb{E}[X^2 + 2XY + Y^2] - (\mu_X + \mu_Y)^2 \\ &= \mathbb{E}[X^2] + 2\,\mathbb{E}[XY] + \mathbb{E}[Y^2] - \mu_X^2 - 2\mu_X\mu_Y - \mu_Y^2 \\ &= \bigl(\mathbb{E}[X^2] - \mu_X^2\bigr) + \bigl(\mathbb{E}[Y^2] - \mu_Y^2\bigr) + 2\bigl(\mathbb{E}[XY] - \mu_X\mu_Y\bigr) \\ &= \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\bigl(\mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y]\bigr). \end{align*} The cross term $\mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y]$ is the covariance of $X$ and $Y$. When $X$ and $Y$ are independent (or merely uncorrelated), $\mathbb{E}[XY] = \mathbb{E}[X]\,\mathbb{E}[Y]$, so the covariance is zero. Therefore $\operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(Y)$. [guided] The computational formula gives $\operatorname{Var}(X + Y) = \mathbb{E}[(X + Y)^2] - (\mathbb{E}[X + Y])^2$. Expanding $(X + Y)^2 = X^2 + 2XY + Y^2$ and using [linearity of expectation](/theorems/1117): \begin{align*} \mathbb{E}[(X+Y)^2] = \mathbb{E}[X^2] + 2\,\mathbb{E}[XY] + \mathbb{E}[Y^2]. \end{align*} Similarly, $(\mathbb{E}[X + Y])^2 = (\mu_X + \mu_Y)^2 = \mu_X^2 + 2\mu_X\mu_Y + \mu_Y^2$. Subtracting and grouping: \begin{align*} \operatorname{Var}(X + Y) = \underbrace{(\mathbb{E}[X^2] - \mu_X^2)}_{\operatorname{Var}(X)} + \underbrace{(\mathbb{E}[Y^2] - \mu_Y^2)}_{\operatorname{Var}(Y)} + 2\underbrace{(\mathbb{E}[XY] - \mu_X\mu_Y)}_{\operatorname{Cov}(X,Y)}. \end{align*} The diagonal terms give $\operatorname{Var}(X) + \operatorname{Var}(Y)$. The cross term is $2\operatorname{Cov}(X, Y)$, which equals $2(\mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y])$. Why does independence kill the covariance? Independence implies $\mathbb{E}[XY] = \mathbb{E}[X]\,\mathbb{E}[Y]$ (the [expectation of a product](/theorems/1120) of independent variables equals the product of expectations). This factorisation is the crucial property — and it holds under the weaker condition of pairwise independence, which is why the theorem statement notes that pairwise independence suffices. [/guided] [/step]