A random variable $X$ is number of boys out of $n$ children. Calculate $\operatorname{Var}(2X-n)$

Question

Let a random variable $X$ be the number of boys out of $n$ children. The probability to have a boy or a girl is $0.5$. Calculate $V(2X-n)$.

I know that $Var(2X-n)=4V(X)$.

$\mathbb{P}(X=k)={1\over 2^n}\binom{n}{k}$. Thus $\mathbb{E}(X)=\sum_{i=1}^n{1\over 2^n}\binom{n}{k}\cdot k$, and $\mathbb{E}(X^2)=\sum_{i=1}^n{1\over 2^n}\binom{n}{k}\cdot k^2$. I'm not sure how to keep on.

Answer 1

$$V(2X - n) = V(2X) = 2^2V(X)$$ $X$ follows a binomial distribution with $n$ trials and $p=0.5$, which has $V(X) = np(1-p)$ $$V(2X - n) = 4np(1-p)$$

Answer 2

HINT $$ \mathrm{Var}\ X = \mathbb{E}\left[(X-\mathbb{E}[X])^2 \right] = \mathbb{E}\left[X^2 \right] - (\mathbb{E}[X])^2 $$

UPDATE

The sums are handled separately. Note that $$ \begin{split} \mathbb{E}[X] &= 2^{-n} \sum_{k=1}^n k \binom{n}{k} \\ &= 2^{-n} \sum_{k=1}^n k \frac{n (n-1)!}{k (k-1)! (n-k)!} \\ &= \frac{n}{2^n} \sum_{k=1}^n \frac{(n-1)!}{(k-1)! (n-k)!} \\ &= \frac{n}{2^n} \sum_{k=0}^{n-1} \frac{(n-1)!}{k! (n-k)!} \\ &= \frac{n}{2^n} 2^{n-1} = n/2. \end{split} $$

The last sum conversion uses the Binomial Theorem. The square sum is transformed similarly to $n/4$.