What is the probability distribution of Total Score if a dice is rolled n-times. Assume dice is fair, and throws are independent. What are these distributions called? What is this branch of mathematics called? Details and background below
More info: 
Playing with Maths after a long time. Basically it started with the Q: What is the Total income distribution as a function of no of coin tosses.
Given that for everytime it's head I get a reward of $R$. The probability of getting a head is $p$. 
In this, I figured out that for $i$ successes in $n$ tosses, the (prob)*(income) is: $p_i R^i = C_{(n,i)}\cdot p^i \cdot (1-p)^{n-i} \cdot R^i$ (binomial dist)
Summing this: $T(n)$ becomes $(p\cdot R + 1-p)^n = (p(R-1) + 1)^n$
Is this expression correct? The trailing + 1 looks a bit odd to me. Is it because the game didn't have any punishment when it's tails, in which case $T(n) = (p\cdot R - (1-p)Q)^n$ where $Q$ is the punishment.
Now I tried to increase the dimensions in the problem(not exactly higher dim version of the same problem, though). How do I solve it(orig prob above)? 
(I know it has  to do with the (i+j+k+l+m+q)th term of expansion $(z_1 + z_2 + z_3 + z_4 + z_5 + z_6)^n$
My main curiosity is: What are these sets of problems? Is there any sub-branch of mathematics dealing with these. Also is there any simpler way of expressing (i,j,k,l,m,q...)th term of (what I hope is )higher dimensional version of Binomial Distribution?
 A: It is indeed a binomial distribution: 
$$P_X(x) = {n \choose x} p^x (1-p)^{n-x} \quad \checkmark$$
Now if your reward amount is multiplied each time you get a good result, then the expected return multiplier is:
$$\begin{align} \operatorname{E}_X[R^X] & = \sum\limits_{x=0}^n P_X(x) R^x \\ & = \sum\limits_{x=0}^n {n\choose x} (pR)^x (1-p)^{n-x} \\ & = (pR + (1-p))^n \\ & = ((R-1)p + 1)^n \quad \checkmark\end{align} \quad \checkmark$$
However if your reward amount is added to each time you get a good result, then the expected return is:
$$\begin{align} \operatorname{E}_X[XR] & = \sum\limits_{x=0}^n P_X(x)\cdot x\cdot R \\ & = R\sum\limits_{x=0}^n {n\choose x} x p^x (1-p)^{n-x} \\ & = n p R\end{align}$$
You do not need binomials to reach this result.  
Let $X_i$ be the event of getting a reward on a single trial.  So $X=\sum_{i=1}^n X_i$ and $\operatorname{E}[X_i]=p$.  Then we have:
$$\begin{align}  \operatorname{E}_X[XR] & = \operatorname{E}[R\sum_{i=1}^n X_i] \\ & = R\sum_{i=1}^n \operatorname{E}[X_i] \\ & = R n p \end{align}$$

When you say "higher dimension version of binomial distribution" for a generating function, I suspect the term you are looking for is: "multinomial distribution".

The expected score of a die rolled $n$ times.
Here we let $T_i \in \{1,2,3,4,5,6\}$ be the score of roll $i$. So $\operatorname{E}[T_i]=\sum\limits_{x=1}^6 \frac x 6 = \frac 7 2$ 
$$\begin{align}\bar T(n) &= \operatorname{E}[\sum\limits_{i=1}^n T_i] \\ & = \sum\limits_{i=1}^n\operatorname{E}[T_i]\\ &= \sum\limits_{i=1}^n \frac 7 2 \\ & = \frac{7n}{2} \end{align}$$
So $\bar T(1) = \frac 7 2, \bar T(2)=7, \ldots, \bar T(100)=35$
This sum of values of multiple dice rolls will have a multinomial distribution, but for moderately large $n$ this will approximate the Normal Distribution. $T\sim\mathcal{N}(\frac {7n}2, \frac{35n}{12})$
