Let $X_1, X_2, ..., X_n$ be a random distribution such that the mean $\mu = 0$ and the variance $\sigma^2$ is unknown.
I'm finding a constant $c$ such that $$U(X) = c \sum^{n-1}_{i = 1}(X_{i+1} - X_i)^2$$ is an unbiased estimator for $\sigma^2$.
So far I've reached $$E[U(X)] = c \sum^{n-1}_{i=1}E[(X_{i+1} - X_i)^2] $$ where $$E[(X_{i + 1} - X_i)^2] $$ $$= E[(X_{i+1})^2 + (X_i)^2 - 2X_{i + 1}X_i] $$ $$ = E[(X_{i+1})^2] + E[(X_i)^2] - 2E[X_{i+1}]E[X_i]$$
which is where I reach a dead end and am not sure how to continue.
Assuming the $X_i$ are independent and identically distributed as $X$, then for all $i$, $\mathbb E[ X_i ] = \mathbb E[X] = \mu$ and $\mathbb E [X_i^2] - \mathbb E[X_i]^2 = \mathbb E [X^2] - \mathbb E[X]^2 = \sigma^2$.
Picking up where you left off and using the above, $$ \mathbb E [ X_{i+1}^2 ] + \mathbb E [ X_{i}^2 ] - 2 \mathbb E [ X_{i+1} ] \mathbb E [ X_i ] = 2 \mathbb E[X^2] - 2 \mathbb E[X]^2 = 2 \sigma^2. $$
Can you take it from here?