Reward for tossing a tail followed by head . A sequence of independent tosses of a biased coin at times $t = 0, 1, 2,...$ On each toss, the probability of a ’head’ is $p$, and the probability of a ’tail’ is $1 − p$. A reward of one unit is given each time that a ’tail’ follows immediately after a ’head.’ Let $R$ be the total reward paid in times $1, 2, ..., n$ . We have to find $E[R]$ and $var(R)$ .
What I did for finding the $E[R]$ :
Let $I_k$ be the reward paid at time $k$ . We have 
$E[I_k]= P(I_k =1) = P(T\ at\ time\ k\ and\ H\ at\ time\ k −1) = p(1−p)$ .
$ E[R] = E[\sum_{k=0}^nI_k] = np(1-p) $
After this I am stuck . I am not able to find the variance . Please help me out .

For the variance, we use the method of indicator random variables that you used to find the mean. We have $R=\sum_{1}^n I_k$. If we can find $E(R^2)$, we will be nearly finished.
Note that $R^2=\sum_1^n I_k^2 +2\sum_{1\le i< j\le n} I_iI_j$. You have already done the calculation of $E(\sum_1^n I_k^2)$, since $I_k^2=I_k$. 
The mixed terms are more complicated. As usual the expectation of the sum is the sum of the expectations, but the expectations are not all equal.
If $j=i+1$, then $I_iI_j=0$. If $j\gt i+1$, then $I_iI_j=1$ with probability $p^2(1-p)^2$. Now it is just a matter of counting. It is more efficient to sum $p^2(1-p)^2$ over all pairs $i\lt j$, and subtract the sum of all $p^2(1-p)^2$ over the $n-1$ consecutive pairs.   
