The chi-squared distribution \(\chi_k^2\) with degree of freedom \(k\) is defined as the following random variable \(Q\)'s distribution: \[Q = \sum_{i=1}^k X_i^2\] where \(\{X_i\} (i = 1, 2, ..., k)\) is a sequence of independent standard normal random variables.
It is simple to prove that (1) \(Q\)'s mean is \(k\) and (2) \(Q\)'s variance is \(2k\); for (1), \begin{align}\mathbb{E}[Q] &= \mathbb{E}\Big[\sum_{i=1}^k X_i^2\Big]\\ &= \sum_{i=1}^k \mathbb{E}[X_i^2] \qquad \text{(linearity of expectation)}\\ &= \sum_{i=1}^k \operatorname{Var}(X_i) \qquad (\text{as }\mathbb{E}[X_i] = 0)\\ &= \sum_{i=1}^k 1\\ &= k.\end{align}
For (2), \begin{align}\operatorname{Var}(Q) &= \operatorname{Var}\Big(\sum_{i=1}^k X_i^2\Big)\\ &= \sum_{i=1}^k \operatorname{Var}(X_i^2) \qquad (\text{as } X_i\text{s are independent})\\ &= \sum_{i=1}^k (\mathbb{E}[X_i^4] - \mathbb{E}[X_i^2]^2)\\ &= \sum_{i=1}^k (\mathbb{E}[X_i^4] - 1) \qquad (\text{as }\mathbb{E}[X_i^2] = \operatorname{Var}(X_i) = 1)\\ &= \sum_{i=1}^k (3 - 1) \qquad (\text{as }\mathbb{E}[X_i^4] = \operatorname{Kurt}(X_i) = 3)\\ &= 2k\end{align} where \(\operatorname{Kurt}(X_i)\) is the kurtosis of \(X_i\).