Find the variance of sum of indicator variables 

Compute $\operatorname{Var} (X)$ where $X = \sum_{i=1}^n x_i$

I am aware of the formula
$$\operatorname{Var} (X) = \sum_{i=1}^9 \operatorname{Var} (X_i) + \sum_{i \ne j} \operatorname{Cov } (X_i, X_j)$$
But I cannot seem to apply it here
Edit:
We know that $\text{Var } (X_i) = 0.234$
Thus $\sum_{i=1}^{9} \text{Var} (X_i) = 9 \times 0.234 = 2.106$
$\displaystyle \sum_{i \ne j} \text{Cov} (X_i, X_j)$, we know that if $j > i + 1$ then $\text{Cov} (X_i, X_j) = 0$, thus $\sum = 9 \times 0.046875 = 0.4219$
Thus $\text{Var} (X) = 9 \times 0.2344 + 0.421875 = 2.53$
 A: \begin{align}
X_i & = \begin{cases} 1 & \text{if the $i$th and $(i+1)$th characters are LW or WL}, \\ 0 & \text{if LL or WW}, \end{cases} \\[10pt]
& = \begin{cases} 1 & \text{with probability } \left(\dfrac 1 4 \times \dfrac 3 4\right) + \left(\dfrac 3 4\times \dfrac 1 4 \right) = \dfrac 3 8, \\[8pt] 0 & \text{with probability } \dfrac 5 8. \end{cases} \\[8pt]
\text{Therefore } \operatorname{var}(X_i) & = \frac 3 8 \times \frac 5 8 = \frac{15}{64}. \\[8pt] {}
\end{align}
\begin{align}
\operatorname{cov}(X_i X_{i+1}) & = \operatorname{E}(X_i X_{i+1}) - (\operatorname{E}X_i)(\operatorname{E}X_{i+1}) \\[8pt]
& = \Pr(X_i X_{i+1} = 1) - \Pr(X_i=1)\Pr(X_{i+1}=1) \\[8pt]
& = \Pr( \text{LWL or WLW)} - \Pr(\text{LW or WL})\Pr(\text{LW or WL)} \\[8pt]
& = \frac 3 {64} + \frac 9 {64} - \left( \frac 3 8 \times \frac 3 8 \right) = \frac {12} {64} - \frac 9 {64} = \frac 3 {64}, \\[10pt]
\text{and } \operatorname{cov}(X_i X_j ) & = 0 \text{ if } |i-j|>1.
\end{align}
\begin{align}
\text{So } \operatorname{var}(X_1+\cdots+X_9) = 9\operatorname{var}(X_1) + 8\times 2\operatorname{cov}(X_1 X_2) = \cdots\cdots
\end{align}
