Generation phi

Okay, consider this:

Define operation ${ O }_{ 1 }({ a }_{ n-1 })=\alpha { a }_{ n-1 }+\beta$ and

${ O }_{ 2 }({ a }_{ n-1 })=\frac { { a }_{ n-1 } }{ \alpha } -\frac { \beta }{ \alpha }$

Observe that ${ O }_{ 1 }$ and ${ O }_{ 2}$ are inverse of each other. The series in the question $S({ a }_{ 0 })$ is formed by repeatedly performing these operations(with equal probability) on the terms.

Now, let's say we did this a few times. Some of the operations we made may be ${ O }_{ 1 }$ and the rest ${ O }_{2}$ . Note that performing ${ O }_{ 1 }$ negates the effect of one ${ O }_{ 2}$ , and vice-versa.

So, if ${ O }_{ 1 }$ is done $p$ times and ${ O }_{ 2}$ , $q$ times while making $2m+1$ terms of $S({a}_{0})$ .

$\Rightarrow p+q=2m+1$

Now we need only bother about which one of $p$ or $q$ is more. let me illustrate: assume $p=5, q=4$ . This means that we performed ${ O }_{ 1 }, 5$ times and ${ O }_{ 2}, 4$ times. Since doing ${ O }_{ 2 }$ $4$ times negates the effect of $4$ ${ O }_{ 1 }$ s we can see that the net effect after performing the $9$ operations is equivalent to performing ${ O }_{ 1 }$ once.

Also note that since $2m+1$ is odd, $p$ and $q$ may differ only by odd numbers.

Hence, to find the expected value, we repeat the following algorithm from $p-q=-(2m+1)$ to $p-q=2m+1$ :

1. Find probability $P(p)$ of $p-q$ taking given value.

2. Multiply $P$ with value of ${a}_{0}$ when ${O}_{1}$ is performed on it $p$ times and ${O}_{2}$ is performed $q$ times.

3. Add to result.

The probability of $p$ being a given value is easy to find and is:

$P(p)=\frac { \left( \begin{matrix} 2m+1 \\ p \end{matrix} \right) }{ { 2 }^{ m+1 } }$

for example, $P(2)$ will give probability of $p$ being $2$ or alternatively, the probability of $q$ being $2m-1$ .

Now what's left to do is to find the general term for when ${O}_{1}$ and ${O}_{2}$ are performed $n$ times.

I am writing the formula here without proof, and leaving it as an exercise to the reader to arrive at this formula.

${ X }_{ n }={ O }_{ 1 }({ O }_{ 1 }(..n\quad times({ a }_{ 0 })))=({ a }_{ 0 }+\frac { \beta }{ \alpha -1 } ){ { \alpha } }^{ n }-\frac { \beta }{ \alpha -1 } \\ { Y }_{ n }={ O }_{ 2 }({ O }_{ 2 }(..n\quad times({ a }_{ 0 })))=({ a }_{ 0 }+\frac { \beta }{ \alpha -1 } )\frac { 1 }{ { \alpha }^{ n } } -\frac { \beta }{ \alpha -1 }$

After performing the algorithm for finding expected value, the required value( $Q$ ) is:

$Q=\frac { 1 }{ { 2 }^{ 2m+1 } } \times \left[ \sum _{ i=1 }^{ m }{ P(m+i){ X }_{ 2i+1 } } +\sum _{ j=1 }^{ m }{ P(m+j){ Y }_{ 2j+1 } } \right]$

$=\frac { 1 }{ { 2 }^{ 2m+1 } } \times \left[ \sum _{ i=1 }^{ m }{ \left( \begin{matrix} 2m+1 \\ m+i \end{matrix} \right) { X }_{ 2i+1 } } +\sum _{ j=1 }^{ m }{ \left( \begin{matrix} 2m+1 \\ m+j \end{matrix} \right) { Y }_{ 2j+1 } } \right]$

After substituting values for ${X}_{i}$ and ${Y}_{j}$ and using binomial theorem, we get:

$Q=\left( { a }_{ 0 }+\frac { \beta }{ \alpha -1 } \right) \frac { { (\alpha +\frac { 1 }{ \alpha } ) }^{ 2m+1 } }{ { 2 }^{ 2m+1 } } -\frac { \beta }{ \alpha -1 }$

Now putting $\alpha =\varphi$ , $\beta =\frac { 1 }{ \varphi }$ , ${a}_{0}=31$ , $2m+1=51$ , we get:

$Q=\frac { { 5 }^{ 25 }\sqrt { 5 } }{ { 2 }^{ 46 } } -1$

Thus, the required answer is $5+25+5+2+46+1=\boxed{84}$

According to the question;

$a_{n}=\varphi a_{n-1} + \frac{1}{\varphi} \quad or \quad a_{n}=\frac{a_{n-1}}{\varphi}-\frac{1}{{\varphi}^2}$

with equal probabilities.

This means if $A_{n}$ is the expected value then

$A_{n}= \frac{1}{2}(\varphi A_{n-1} + \frac{1}{\varphi}) +\frac 1 2 (\frac{A_{n-1}}{\varphi}-\frac{1}{{\varphi}^2})$

$\Rightarrow A_{n}= \frac{1}{2}(\varphi A_{n-1} + \frac{1}{\varphi} + \frac{A_{n-1}}{\varphi}-\frac{1}{{\varphi}^2})$

$\Rightarrow A_{n}= \frac{1}{2}(A_{n-1}(\varphi +\frac{1}{\varphi}) + (\frac{1}{\varphi}-\frac{1}{{\varphi}^2}))$

Putting the value of $\varphi=\frac{\sqrt{5}+1}{2}$ and $\frac{1}{\varphi}=\frac{\sqrt{5}-1}{2}$ we get ,

$\Rightarrow A_{n}= \frac{1}{2}(A_{n-1}\sqrt{5} + \sqrt{5}-2)$

$\Rightarrow A_{n}= \frac{1}{2}(\sqrt{5}(A_{n-1} + 1) -2)$

$\Rightarrow A_{n}= \frac{1}{2}(\sqrt{5}(A_{n-1} + 1)) -1$

$\Rightarrow A_{n} + 1= \frac{\sqrt{5}}{2}(A_{n-1} + 1)$

Now let $A_{n}+1=T_{n}$ with $T_0=32$

$\Rightarrow T_{n} = \frac{\sqrt{5}}{2}T_{n-1}$

Now a GP if formed with common ratio $\frac{\sqrt{5}}{2}$

This gives us $T_{51}=T_{0} (\frac{\sqrt{5}}{2})^{51}$

and $A_{51}=T_{51}-1=(32) (\frac{\sqrt{5}}{2})^{51}-1$

$\Rightarrow A_{51}=\frac{5^{25} \sqrt{5}}{2^{46}}-1$

$\Rightarrow a+b+c+d+e+f=5+25+5+2+46+1=\boxed{84}$