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user103828's answer dated Aug 28 '17 that uses recursion to define $p_{n}$ and $q_{n}$, is very light on details and makes it sound as if you can directly solve of $p_{n} = q_{n}$ from the two equations.

Note that the two equations

$$p_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}q_{n-1}$$

don't offer independent constraints on the variables since $p_{n} = 1 - q_{n}$. So subtract the equations $p_{n} - q_{n}$:

$$p_{n} - q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1} - \frac{1}{n} - \frac{n - 2}{n}q_{n-1}$$ $$p_{n} - q_{n} = \frac{n - 2}{n}(p_{n-1} - q_{n-1})$$

At this point we can state for the base case, $n=2$, we know $p_{2} = q_{2} = \frac{1}{2}$. Now if we assume $q_{n-1} = p_{n-1}$, clearly the right-hand side of our last equation becomes zero, and $q_{n} = p_{n}$.

Since $q_{n} + p_{n} = 1$, then $\boxed{p_{n} = q_{n} = \frac{1}{2}}$.

It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_{n}$ or $q_{n}$ even though this is how the aforementioned user103828's answer makes it sound.

The other answer that uses recursion to define $p_{n}$ and $q_{n}$ is very light on details and makes it sound as if you can directly solve of $p_{n} = q_{n}$ from the two equations.

Note that the two equations

$$p_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}q_{n-1}$$

don't offer independent constraints on the variables since $p_{n} = 1 - q_{n}$. So subtract the equations $p_{n} - q_{n}$:

$$p_{n} - q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1} - \frac{1}{n} - \frac{n - 2}{n}q_{n-1}$$ $$p_{n} - q_{n} = \frac{n - 2}{n}(p_{n-1} - q_{n-1})$$

At this point we can state for the base case, $n=2$, we know $p_{2} = q_{2} = \frac{1}{2}$. Now if we assume $q_{n-1} = p_{n-1}$, clearly the right-hand side of our last equation becomes zero, and $q_{n} = p_{n}$.

Since $q_{n} + p_{n} = 1$, then $\boxed{p_{n} = q_{n} = \frac{1}{2}}$.

It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_{n}$ or $q_{n}$ even though this is how the other solution makes it sound.

The other answer that uses recursion to define $p_{n}$ and $q_{n}$ is very light on details and makes it sound as if you can directly solve of $p_{n} = q_{n}$ from the two equations.

Note that the two equations

$$p_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}q_{n-1}$$

don't offer independent constraints on the variables since $p_{n} = 1 - q_{n}$. So subtract the equations $p_{n} - q_{n}$:

$$p_{n} - q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1} - \frac{1}{n} - \frac{n - 2}{n}q_{n-1}$$ $$p_{n} - q_{n} = \frac{n - 2}{n}(p_{n-1} - q_{n-1})$$

At this point we can state for the base case, $n=2$, we know $p_{2} = q_{2} = \frac{1}{2}$. Now if we assume $q_{n-1} = p_{n-1}$, clearly the right-hand side of our last equation becomes zero, and $q_{n} = p_{n}$.

Since $q_{n} + p_{n} = 1$, then $\boxed{p_{n} = q_{n} = \frac{1}{2}}$.

It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_{n}$ or $q_{n}$ even though this is how the other solution makes it sound.

user103828's answer dated Aug 28 '17 that uses recursion to define $p_{n}$ and $q_{n}$, is very light on details and makes it sound as if you can directly solve of $p_{n} = q_{n}$ from the two equations.

Note that the two equations

$$p_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}q_{n-1}$$

don't offer independent constraints on the variables since $p_{n} = 1 - q_{n}$. So subtract the equations $p_{n} - q_{n}$:

$$p_{n} - q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1} - \frac{1}{n} - \frac{n - 2}{n}q_{n-1}$$ $$p_{n} - q_{n} = \frac{n - 2}{n}(p_{n-1} - q_{n-1})$$

At this point we can state for the base case, $n=2$, we know $p_{2} = q_{2} = \frac{1}{2}$. Now if we assume $q_{n-1} = p_{n-1}$, clearly the right-hand side of our last equation becomes zero, and $q_{n} = p_{n}$.

Since $q_{n} + p_{n} = 1$, then $\boxed{p_{n} = q_{n} = \frac{1}{2}}$.

It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_{n}$ or $q_{n}$ even though this is how the aforementioned user103828's answer makes it sound.

The other answer that uses recursion to define $p_{n}$ and $q_{n}$ is very light on details and makes it sound as if you can directly solve of $p_{n} = q_{n}$ from the two equations.

Note that the two equations

$$p_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}q_{n-1}$$

don't offer independent constraints on the variables since $p_{n} = 1 - q_{n}$. So just working with the equations for $p_{n}$:

$$1 - q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = 1 - \frac{1}{n} - \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{n - 1}{n} - \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n} - \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}[1 - p_{n-1}]$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}[1 - p_{n-1}] + \frac{n - 2}{n}p_{n - 1} - \frac{n - 2}{n}p_{n - 1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n - 1} + \frac{n - 2}{n}[1 - 2p_{n-1}]$$ $$q_{n} = p_{n} + \frac{n - 2}{n}[1 - 2p_{n-1}]$$ $$q_{n} - p_{n} = \frac{n - 2}{n}[1 - p_{n-1} - p_{n-1}]$$ $$q_{n} - p_{n} = \frac{n - 2}{n}[q_{n-1} - p_{n-1}]$$

At this point we can state for the base case, $n=2$, we know $p_{n} = q_{n} = \frac{1}{2}$. Now if we assume $q_{n-1} = p_{n-1}$, clearly the right-hand side of our last equation becomes zero, and $q_{n} = p_{n}$.

Since $q_{n} + p_{n} = 1$, then $\boxed{p_{n} = q_{n} = \frac{1}{2}}$.

It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_{n}$ or $q_{n}$ even though this is how the other solution makes it sound.

The other answer that uses recursion to define $p_{n}$ and $q_{n}$ is very light on details and makes it sound as if you can directly solve of $p_{n} = q_{n}$ from the two equations.

Note that the two equations

$$p_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1}$$ $$q_{n} = \frac{1}{n} + \frac{n - 2}{n}q_{n-1}$$

don't offer independent constraints on the variables since $p_{n} = 1 - q_{n}$. So subtract the equations $p_{n} - q_{n}$:

$$p_{n} - q_{n} = \frac{1}{n} + \frac{n - 2}{n}p_{n-1} - \frac{1}{n} - \frac{n - 2}{n}q_{n-1}$$ $$p_{n} - q_{n} = \frac{n - 2}{n}(p_{n-1} - q_{n-1})$$

At this point we can state for the base case, $n=2$, we know $p_{2} = q_{2} = \frac{1}{2}$. Now if we assume $q_{n-1} = p_{n-1}$, clearly the right-hand side of our last equation becomes zero, and $q_{n} = p_{n}$.

Since $q_{n} + p_{n} = 1$, then $\boxed{p_{n} = q_{n} = \frac{1}{2}}$.

It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_{n}$ or $q_{n}$ even though this is how the other solution makes it sound.