How do you calculate the expected value of geometric distribution without diffrentiation? Is there any way I can calculate the expected value of geometric distribution without diffrentiation? All other ways I saw here have diffrentiation in them.
Thanks in advance!

The problem can be viewed in a different perspective to understand more intuitively. Let's see the following definition.
"A person tosses a coin, if head comes he stops, else he passes the coin to the next person. For the next person, he follows the same process: If head comes he stops, else passes the coin to next person and so on."
So the process can be modelled as,
$$ 
X = 
\begin{cases}
1, & \text{if $head$ occurs} \\
1 + Y, & \text{if $tail$ occurs}
\end{cases}
$$
where $Y$ denotes the next person.
Similarly for $Y$, the equation can be written as,
$$ 
Y = 
\begin{cases}
1, & \text{if $head$ occurs} \\
1 + Y', & \text{if $tail$ occurs}
\end{cases}
$$
where $Y'$ is the next after $Y$ and so on.
We see that there is no difference between $X$ and $Y$, both toss the coin and stop if head comes else pass the coin to next person who does the same. If both the events were occurring separately and independently (i.e. $Y$ had a coin already and didn't know it came from $X$), their average (or expected) values would be same.
Let's calculate average value of $X$ using the formula of expected values.
Here $p$ is the probability of head and $q$ is the probability of tail.
$$
E(X) = p \times 1 + q \times (1 + Y) \tag{1}\label{eq1}
$$
Let's see the formula in an intuitive way. Suppose we ask $X$, What is the average number of tosses that you need to get heads?. He will be like, My expected value is 1 into p if head occurs in the first toss else (1 + no. of tosses that $Y$ had to make) into q. And $Y$ is like so on an average I had to do $E(Y)$ tosses to get heads. So $X$'s final statement is 1 into p if head comes in first toss else 1 + $E(Y)$ tosses tail comes.
So equation can be reduced as,
$$
\begin{align}
&E(X) = p \times 1 + q \times (1 + E(Y)) && \text{} \tag{2}\label{eq2} \\
\Rightarrow \ &E(X) = p \times 1 + q \times (1 + E(X)) && \text{Since $E(X) = E(Y)$} \\
\Rightarrow \ &E(X) = 1/p
\end{align}
$$

Note that, $\eqref{eq1}$ is very different from $\eqref{eq2}$. 
$(2)$ can be solved as seen above, but $(1)$ just continues to unfold further like $E(X) = 1 \times p + q \times (1 + 1) \times p + q \times (1 + Y') \times q$ and so on. Although this can also be reduced to $E(X) = 1 \times p + q \times (1 + E(X))$, but I wanted to follow the process intuitively.

Since it can seem off, to just put the average value of $Y$ in $(1)$, improvements with strong reasoning are welcomed.
