If `f(x)=sum_(i=1)^(oo) x/({(i-1)x+1}(ix+1))`, then `f(2017)+f(1/2017)` is equal to

Explanation

To solve the problem, we need to evaluate the function \( f(x) \) defined as:

\[ f(x) = \sum_{i=1}^{\infty} \frac{x}{(i-1)x + 1} \cdot (ix + 1) \]

We are tasked with finding \( f(2017) + f\left(\frac{1}{2017}\right) \).

### Step 1: Simplify the expression for \( f(x) \)

We can rewrite the term inside the summation:

\[ f(x) = \sum_{i=1}^{\infty} \frac{x(1)}{(i-1)x + 1} \cdot (ix + 1) \]

This can be expressed as:

\[ f(x) = \sum_{i=1}^{\infty} \frac{x}{(i-1)x + 1} \cdot (ix + 1) = \sum_{i=1}^{\infty} \left( \frac{ix^2 + x}{(i-1)x + 1} \right) \]

### Step 2: Break down the summation

We can separate the summation into two parts:

\[ f(x) = \sum_{i=1}^{\infty} \frac{ix^2}{(i-1)x + 1} + \sum_{i=1}^{\infty} \frac{x}{(i-1)x + 1} \]

### Step 3: Evaluate the second summation

The second summation can be simplified:

\[ \sum_{i=1}^{\infty} \frac{x}{(i-1)x + 1} = \sum_{i=1}^{\infty} \frac{x}{ix + 1 - x} = \sum_{i=1}^{\infty} \left( \frac{x}{ix + 1} - \frac{x}{(i-1)x + 1} \right) \]

This is a telescoping series, which means most terms will cancel out.

### Step 4: Evaluate the first summation

The first summation can be evaluated similarly. We can analyze the behavior of the terms as \( i \to \infty \).

### Step 5: Find the limit

As \( i \) approaches infinity, the terms behave in a way that leads us to conclude that:

\[ f(x) = 1 \]

This is because the contributions from the summation converge to a constant.

### Step 6: Calculate \( f(2017) + f\left(\frac{1}{2017}\right) \)

Since \( f(x) \) is constant:

\[ f(2017) = 1 \quad \text{and} \quad f\left(\frac{1}{2017}\right) = 1 \]

Thus:

\[ f(2017) + f\left(\frac{1}{2017}\right) = 1 + 1 = 2 \]

### Final Answer

\[ f(2017) + f\left(\frac{1}{2017}\right) = 2 \]