Let $(W_t)_{t\ge0}$ be a Brownian motion and consider the random variables $$Y_n=\sum_{i=1}^{2^n}|W_{i2^{-n}}-W_{(i-1)2^{-n}}|\text{ where }n\in\mathbb{N}.$$ I want to compute the expectation of one such random variable. I know that from linearity of the expectation we can get the expectation inside the sum $$\mathbb{E}Y_n=\sum_{i=1}^{2^n}\mathbb{E}|W_{i2^{-n}}-W_{(i-1)2^{-n}}|.$$ I also think that the increments are such that $W_{i2^{-n}}-W_{(i-1)2^{-n}}\sim\mathcal{N}(0,2^{-n})$ from the definition of a Brownian motion.
However I don't know how to continue. I should obtain $\mathbb{E}Y_n = 2^{n/2}\mathbb{E}|W_1$|.
Recall that if $X \sim \mathcal{N}(0,\sigma^2)$ then for $t \geq 0$ $$P(|X|\leq t)=P(-t\leq X\leq t)=\int_{[-t,t]}f_X(x)dx=F_X(t)-F_X(-t) $$ which implies $f_{|X|}(t)=2f_X(t)$. Therefore $$E[|X|]=2\int_{[0,\infty)}t\,f_{X}(t)dt=-2\sigma^2\int_{[0,\infty)}\underbrace{-\frac{t}{\sigma^2}f_X(t)}_{=f'_X(t)}dt=2\sigma^2\frac{1}{\sqrt{2\pi \sigma^2}}=\sigma\sqrt{\frac{2}{\pi}}$$ We have $W_{i2^{-n}}-W_{(i-1){2^{-n}}}\sim \mathcal{N}(0,2^{-n})$ so $$E[Y_n]=\sqrt{\frac{2}{\pi}}\sum_{i=1}^{2^{n}}2^{-\frac{n}{2}}=E[|W_1|]2^{-\frac{n}{2}}2^n=E[|W_1|]2^{\frac{n}{2}}$$