Assume n independent experiments with probabilities of $p_1,...,p_n$. Let's define the random variable as this:
$$X= \begin{cases} \ 1, & \text{if exactly 1 of n experiments was successful} \\2, & \text{if exactly 2 of n experiments were successful} \\\ ... \\\\ n,& \text{if exactly n of n experiments were successful} \end{cases}$$ Now the expected value (mean) of X would be: $$E[X]=\sum_{i=1}^n X_iP(X_i)$$ This expected value would be equal to sum of $p_1,...,p_n$. In other words: $$\sum_{i=1}^n X_iP(X_i)=\sum_{i=1}^n p_i $$ Why the expected value is equal to sum of $p_1,...,p_n$? I'd be grateful if someone writes the mathematical proof coherently so I can use it in a scientific research.
It would be better to write "$P[X=i]$" rather than "$P[X_i]$," especially since we do not know what "$X_i$" is supposed to mean (if it is inside the $P[\cdot]$ function then it must be an event). So, $$E[X] = \sum_{i=0}^n iP[X=i]$$
You can derive the desired result using indicator functions together with the fact that the expectation of a sum is the sum of expectations: Define $$ Y_i = \left\{ \begin{array}{ll} 1 &\mbox{ if experiment $i$ is successful} \\ 0 & \mbox{ otherwise} \end{array} \right. $$ Then $$X = Y_1 + Y_2 + ... + Y_n \implies E[X]=E[Y_1] + E[Y_2] + ... + E[Y_n]$$ But of course you can compute $E[Y_i]$ for each $i\in\{1, \ldots, n\}$ to get the desired result. Also, that result is true regardless of whether or not the experiments are independent.