Statement-1: `sum_(r=0)^(n) (1)/(r+1) ""^(n)C_(r) = (1)/((n+1)x) {( 1 + x)^(n+1) -1}^(-1)`
Statement-2: ` sum_(r=0)^(n) (""^(n)C_(r))/(r+1) = (2^(n+1))/(n+1)`.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given problem, we need to analyze both statements and verify their validity step by step.

### Step 1: Analyze Statement 1
We start with the first statement:
\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{(n+1)x} \left( (1+x)^{n+1} - 1 \right)^{-1} \]

#### Step 1.1: Rewrite the left-hand side
The term \(\binom{n}{r}\) can be expressed as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Thus, we can rewrite the left-hand side:
\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \sum_{r=0}^{n} \frac{n!}{(r+1)!(n-r)!} \]
This can be simplified by noting that:
\[ \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \binom{n+1}{r+1} \]
Therefore, we can rewrite the sum as:
\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} \sum_{r=1}^{n+1} \binom{n+1}{r} \]

#### Step 1.2: Evaluate the sum
The sum \(\sum_{r=1}^{n+1} \binom{n+1}{r}\) equals \(2^{n+1} - 1\) (since it includes all terms except for \(r=0\)). Thus, we have:
\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} (2^{n+1} - 1) \]

### Step 2: Compare with the right-hand side
Now we need to check if this matches the right-hand side:
\[ \frac{1}{(n+1)x} \left( (1+x)^{n+1} - 1 \right)^{-1} \]
Substituting \(x = 1\):
\[ \frac{1}{(n+1)} \left( 2^{n+1} - 1 \right) \]
This confirms that Statement 1 is true.

### Step 3: Analyze Statement 2
Now we check the second statement:
\[ \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} = \frac{2^{n+1}}{n+1} \]

#### Step 3.1: Use the result from Statement 1
From our previous result, we found:
\[ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{2^{n+1} - 1}{n+1} \]
This means that:
\[ \sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} = \frac{2^{n+1} - 1}{n+1} \]

#### Step 3.2: Compare with the right-hand side
The right-hand side of Statement 2 is:
\[ \frac{2^{n+1}}{n+1} \]
Since \(\frac{2^{n+1} - 1}{n+1} \neq \frac{2^{n+1}}{n+1}\), we conclude that Statement 2 is false.

### Conclusion
- Statement 1 is **True**.
- Statement 2 is **False**.