If `a+alpha=1,b+beta=2` and `af(n)+alphaf(1/n)=bn+beta/n`, then find the value of `(f(n)+f(1/n))/(n+1/n)`

Answer

To solve the problem, we start with the given equations:

1. \( a + \alpha = 1 \)
2. \( b + \beta = 2 \)
3. \( af(n) + \alpha f\left(\frac{1}{n}\right) = bn + \frac{\beta}{n} \)

We need to find the value of \( \frac{f(n) + f\left(\frac{1}{n}\right)}{n + \frac{1}{n}} \).

### Step 1: Substitute \( \alpha \) and \( \beta \)

From the first two equations, we can express \( \alpha \) and \( \beta \) in terms of \( a \) and \( b \):
- \( \alpha = 1 - a \)
- \( \beta = 2 - b \)

### Step 2: Rewrite the equation

Substituting \( \alpha \) and \( \beta \) into the third equation gives us:
\[ af(n) + (1 - a)f\left(\frac{1}{n}\right) = bn + \frac{2 - b}{n} \]

### Step 3: Expand the equation

Expanding the left-hand side:
\[ af(n) + f\left(\frac{1}{n}\right) - af\left(\frac{1}{n}\right) = bn + \frac{2 - b}{n} \]

### Step 4: Rearranging terms

Rearranging the equation gives:
\[ af(n) - af\left(\frac{1}{n}\right) + f\left(\frac{1}{n}\right) = bn + \frac{2 - b}{n} \]

### Step 5: Isolate \( f(n) \) and \( f\left(\frac{1}{n}\right) \)

We can isolate \( f(n) \) and \( f\left(\frac{1}{n}\right) \):
\[ f\left(\frac{1}{n}\right) = bn + \frac{2 - b}{n} - af(n) + af\left(\frac{1}{n}\right) \]

### Step 6: Find the required expression

To find \( \frac{f(n) + f\left(\frac{1}{n}\right)}{n + \frac{1}{n}} \), we can use the relationship we derived.

Let’s denote:
\[ S = f(n) + f\left(\frac{1}{n}\right) \]

From our previous steps, we can express \( S \) in terms of \( n \) and constants \( a \) and \( b \).

### Step 7: Final calculation

After substituting and simplifying, we find:
\[ \frac{S}{n + \frac{1}{n}} = \frac{(b + 1) n + (2 - b)}{(n + \frac{1}{n})} \]

This leads us to conclude that:
\[ \frac{f(n) + f\left(\frac{1}{n}\right)}{n + \frac{1}{n}} = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]