Hello I am just baffled on what to do to derive or begin this question.
My futile attempt was to let N approach infinity (b/c after infinite trials the mean should reach to Np under assumption) and see if there were any patterns or cancelations that went on to get rid of some parameters.
How would you start the proof?
The mean-value is given by $$<k>=\frac{\sum_{k=0}^{n} k {n \choose k} p^k q^{n-k}}{(p+q)^n}~~~~(1)$$
By binomial theorem we have $$(p+q)^n = \sum_{k=0}^{n} {n \choose k} p^k q^{n-k}.$$ Differentiate this identity partially w. r. t. $p$ to get $$n(p+q)^{n-1}=\sum_{k=0}^{n} k {n \choose k} p^{k-1} q^{n-k}$$. Multiplying both the sides by $p$, we get $$\sum_{k=0}^{n} k {n\choose k} p^k q^{n-k}=np(p+q)^{n-1}~~~~(2)$$ Putting this in (1) we get $$<k>=\frac{np}{p+q}$$. Finally by letting $p+q=1$ we get $<k>=np.$
For $i=1,2,\dots,n$ let $X_i$ be a random variable whose value is $1$ when toss $i$ is heads and $0$ when it is tails. We want $E\left(\sum_{i=1}^n X_i\right)$.
What do you know about the expectation of a sum?
The expectation of a sum is the sum of the expectations:$$E\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^nE(X_i)=nE(X_1),$$ since all the $X_i$ have the same distribution. Now $$E(X_1)=p\cdot1+(1-p)\cdot0=p,$$ and so $$E\left(\sum_{i=1}^n X_i\right)=np$$
