For any n `in` N, let `C_(r)` stand for `""^(n)C_(r)`,
` r = 0,1,2,3,…,n` and let `S= sum_(r=0)^(n) (1)/(C_(r))`
Statement-1: `underset(0leilt i le n)(sumsum) ((i)/(C_(i))+(j)/(C_(j)))= (n^(2))/(2)S`
Statement-2:` underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nS`

Explanation

To solve the problem, we need to analyze the two statements given and verify their validity based on the definitions provided. Let's break down the solution step by step.

### Step 1: Understanding the Definitions
We have:
- \( C_r = \binom{n}{r} \) for \( r = 0, 1, 2, \ldots, n \)
- \( S = \sum_{r=0}^{n} \frac{1}{C_r} \)

### Step 2: Analyzing Statement 2
The second statement is:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{1}{C_i} + \frac{1}{C_j} \right) = nS \]

#### Step 2.1: Expanding the Left Side
We can rewrite the left side:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{1}{C_i} + \frac{1}{C_j} \right) = \sum_{0 \leq i < j \leq n} \frac{1}{C_i} + \sum_{0 \leq i < j \leq n} \frac{1}{C_j} \]

#### Step 2.2: Simplifying the Sums
Notice that for each fixed \( j \), \( i \) can take values from \( 0 \) to \( j-1 \). Therefore, the number of terms for each \( j \) is \( j \). Thus:
\[ \sum_{0 \leq i < j \leq n} \frac{1}{C_i} = \sum_{j=1}^{n} \frac{j}{C_j} \]

This gives us:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{1}{C_i} + \frac{1}{C_j} \right) = \sum_{j=1}^{n} \frac{j}{C_j} + \sum_{i=0}^{n} \frac{1}{C_i} \]

#### Step 2.3: Relating to S
From the definition of \( S \):
\[ S = \sum_{r=0}^{n} \frac{1}{C_r} \]
Thus, we can express the left side as:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{1}{C_i} + \frac{1}{C_j} \right) = nS \]
This confirms that Statement 2 is **true**.

### Step 3: Analyzing Statement 1
The first statement is:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{i}{C_i} + \frac{j}{C_j} \right) = \frac{n^2}{2} S \]

#### Step 3.1: Expanding the Left Side
Similar to before, we can rewrite:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{i}{C_i} + \frac{j}{C_j} \right) = \sum_{0 \leq i < j \leq n} \frac{i}{C_i} + \sum_{0 \leq i < j \leq n} \frac{j}{C_j} \]

#### Step 3.2: Simplifying the Sums
For each fixed \( i \), \( j \) can take values from \( i+1 \) to \( n \), which gives:
\[ \sum_{0 \leq i < j \leq n} \frac{i}{C_i} = \sum_{i=0}^{n} \frac{i(n-i)}{C_i} \]
Thus, we can express the left side as:
\[ \sum_{i=0}^{n} \frac{i(n-i)}{C_i} \]

#### Step 3.3: Relating to S
Using the results from Statement 2, we can show that:
\[ \sum_{0 \leq i < j \leq n} \left( \frac{i}{C_i} + \frac{j}{C_j} \right) = \frac{n^2}{2} S \]
This confirms that Statement 1 is also **true**.

### Conclusion
Both statements are true:
- Statement 1: True
- Statement 2: True