Update 5-1.md

As we want to compute the variance of V_n, it will be decomposed to X_1, \cdots X_n instead of V_1, \cdots, V_n, so this typo is actually quite apparent.

docs/Chap05/Problems/5-1.md

@@ -82,7 +82,7 @@ Why is the second way more rigorous than the first? Both ways condition on the v
 
 **b.** Defining $V_n$ and $X_j$ as in part (a), we want to compute $\text{Var}[V_n]$, where $n_i = 100i$. The $X_j$ are pairwise independent, and so by equation $\text{(C.29)}$,
 
-$$\text{Var[$V_1$]} + \text{Var[$V_2$]} + \cdots + \text{Var[$V_n$]}.$$
+$$\text{Var[$V_n$]} = \text{Var[$X_1$]} + \text{Var[$X_2$]} + \cdots + \text{Var[$X_n$]}.$$
 
 Since $n_i = 100i$, we see that $n_{i + 1} - n_i = 100(i + 1) - 100i = 100$. Therefore, with probability $99 / 100$, the increase in the value represented by the counter due to the $j$th $\text{INCREMENT}$ operation is $0$, and with probability $1 / 100$, the value represented increases by $100$. Thus, by equation $\text{(C.27)}$,