No, $\sum kP\left (k \right )$ doesn't generalize to fibonacci numbers,

for example consider the followimg distribution:

this is a valid distribution as $\sum P(k) = 1$,

but $\sum kP\left (k \right )$ isn't giving any fibonacci number here, or it may be somehow which i cann't see..

$\sum kP\left (k \right )$ is the definition of expectation of a discrete random variable.

**let me explain above solution more thoroughly,**

**step 1:** decide your random variable:

let 'k' be a random variable which denotes the no of times you need to pick a ball from bag.

**step 2:** define your distribution..

**k=1, ** it means you get white ball in first trial only

$p(k=1) = \frac{1}{2}$

**k=2, you don't get a white in first go, but you get it in second one,**

$p(k=2) = p(not getting white)\times p(getting white) = \frac{1}{2}\times \frac{1}{2}$

$p(k=2) =\left ( \frac{1}{2} \right )^{2}$

**k=3, you get white in 3rd attempt(or trial):**

**$p(k=3) = p(not getting white)\times p(not getting white)\times p(getting white) = \frac{1}{2}\times \frac{1}{2} \times \frac{1}{2}$**

$p(k=3) =\left ( \frac{1}{2} \right )^{3}$

and so on..

so our distribution looks like this:

k |
1 |
2 |
3 |
4 |
.. |

p(k) |
$\left ( \frac{1}{2} \right )^{1}$ |
$\left ( \frac{1}{2} \right )^{2}$ |
$\left ( \frac{1}{2} \right )^{3}$ |
$\left ( \frac{1}{2} \right )^{4}$ |
.. |

**step3:** **we have to conform that this is a valid random distribution,**

to do so, we have to prove $\sum p(k) = 1$.

$\sum p\left (k \right ) = \left ( \frac{1}{2} \right )^{1} + \left ( \frac{1}{2} \right )^{2} + \left ( \frac{1}{2} \right )^{3} + \left ( \frac{1}{2} \right )^{4} + ...$

$=\frac{ \frac{1}{2}}{1 - \frac{1}{2}} = 1$

Therefore this is a valid distribution.

**step4: Calculate expected value of random variable**

**$E\left ( k \right ) = \sum k\cdot p\left ( k \right )$**

$E = E\left ( k \right ) = 1\times\left ( \frac{1}{2} \right )^{1} + 2\times\left ( \frac{1}{2} \right )^{2} + 3\times\left ( \frac{1}{2} \right )^{3} + ...$ --(I)

Now, all we have to do is to calculate the sum of this sequence,

these's many ways to solve this sequence, one way is how i did it..

Multiply above equation by half

$\frac{E}{2} =\frac{1}{2}\times \left (1\times\left ( \frac{1}{2} \right )^{1} + 2\times\left ( \frac{1}{2} \right )^{2} + 3\times\left ( \frac{1}{2} \right )^{3} + ... \right )$

$\frac{E}{2}\ =1\times\left ( \frac{1}{2} \right )^{2} + 2\times\left ( \frac{1}{2} \right )^{3} + 3\times\left ( \frac{1}{2} \right )^{4} + ...$ --(II)

subtract eq (II) from eq (I) as follows ( **eq (II) - eq(I)** ):

$E=1\times\left ( \frac{1}{2} \right )^{1} + 2\times\left ( \frac{1}{2} \right )^{2} + 3\times\left ( \frac{1}{2} \right )^{3} +4\times\left ( \frac{1}{2} \right )^{4} + ...$

$- \frac{E}{2}\ =\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\times\left ( \frac{1}{2} \right )^{2} + 2\times\left ( \frac{1}{2} \right )^{3} + 3\times\left ( \frac{1}{2} \right )^{4} + ...$

-----------------------------------------------------------------------------------------------------------------

$E - \frac{E}{2}\ =\ \ \ \ \ \ \left (\frac{1}{2}\right )^{1} + \ \ \ \ \ \left ( \frac{1}{2} \right )^{2} +\ \ \ \ \ \ \ \left ( \frac{1}{2} \right )^{3} +\ \ \ \ \ \ \ \left ( \frac{1}{2} \right )^{4} + ...$

which gives,

$\frac{E}{2}=\left (\frac{1}{2} \right )^{1} + \left (\frac{1}{2} \right )^{2} + \left (\frac{1}{2} \right )^{3} + \left (\frac{1}{2} \right )^{4} + \ ....$

**Now this is GP, use infinite sum formula**

$\frac{E}{2} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$

$E = 2$

This is it. No need to use fibonacci formula

I should mention there are lot of ways to solve this same problem