Mean average of exponential functions I'm new to math stack exchange. Hope this goes well!
So, I have a set of exponential functions each defined in the form
\begin{aligned}
f_i(x)=a_ie^{-b_ix}+c_i
\end{aligned}
where $(a_i, b_i, c_i) \in \Bbb R^+$. (EDIT: Changed from $\Bbb R$ to $\Bbb R^+$)
Is it correct to say that:
\begin{aligned}
\frac{1}{n}\sum_{i=1}^{n}f_i(x) &= (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i
\end{aligned}
for $n \in \Bbb Z^+$?
If so, how can we prove it?
If not, why not?
 A: You wrote :

\begin{aligned}
f_i(x)=a_ie^{-b_ix}+c_i
\end{aligned}
  where $(a_i, b_i, c_i) \in \Bbb R^+$. 
  Is it correct to say that:
  \begin{aligned}
\frac{1}{n}\sum_{i=0}^{n}f_i(x) &= (\frac{1}{n}\sum_{i=0}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=0}^{n}b_i)x}+\frac{1}{n}\sum_{i=0}^{n}c_i
\end{aligned}

First there is a mistake : A sum from $i=0$ to $i=n$ counts $n+1$ terms, not $n$. 
So, for average, you should write $\frac{1}{n+1}\sum_{i=0}^{n}$ or $\frac{1}{n}\sum_{i=1}^{n}$ ,but not $\frac{1}{n}\sum_{i=0}^{n}$ .
I suppose that it is a typo and that the proposed relationship is 
$$\frac{1}{n}\sum_{i=1}^{n}(a_ie^{-b_ix}+c_i)
 = (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i$$
It is easy to find a lot of counter-examples . 
For example, check your formula with 
$$n=2\:;\:a_1=a_2=1\:;\:b_1=1\:;\:b_2=3\:;\:c _1=c_2=0$$
$$\begin{cases}
\frac{1}{n}\sum_{i=1}^{n}(a_ie^{-b_ix}+c_i) =\frac12 (e^{-x}+e^{-3x})\\
(\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i =e^{-2x}
\end{cases}$$
$$\frac12 (e^{-x}+e^{-3x})\neq e^{-2x}$$
Thus  $\quad \boxed{\frac{1}{n}\sum_{i=1}^{n}f_i(x) = (\frac{1}{n}\sum_{i=1}^{n}a_i)e^{-(\frac{1}{n}\sum_{i=1}^{n}b_i)x}+\frac{1}{n}\sum_{i=1}^{n}c_i\quad \text{is false.}}$
Another obvious counter-example, with   $a_k=1\:;\:b_k=k\:;\: c_k=0$ :
$$\quad \frac{1}{n}\sum_{k=0}^{n}e^{-kx} \neq e^{-\frac{n+1}{2}x}$$
because $\frac{1}{n}\sum_{k=0}^{n}e^{-kx}=\frac{e^{-x}-e^{-(n+1)x}}{n(1-e^{-x})}\neq e^{-\frac{n+1}{2}x}$ .
Note that the proposed relationship is true in the trivial case $b_1=b_2=…=b_n$ .
