Statement-1: `(C_(0))/(2.3)- (C_(1))/(3.4) +(C_(2))/(4.5)-.............+............+(-1)^(n) (C_(n))/((n+2)(n+3))= (1)/((n+1)(n+2))`
Statement-2:`(C_(0))/(k)- (C_(1))/(k+1) +(C_(2))/(k+3)+............+(-1)^(n) (C_(n))/(k+n)=int_(0)^(1)x^(k-1) (1 - x)^(n) dx`

Options

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

Explanation

To solve the given problem, we will analyze both statements step by step.

### Step 1: Understanding Statement 1
The first statement is:
\[ \frac{C_0}{2 \cdot 3} - \frac{C_1}{3 \cdot 4} + \frac{C_2}{4 \cdot 5} - \ldots + (-1)^n \frac{C_n}{(n+2)(n+3)} = \frac{1}{(n+1)(n+2)} \]
where \( C_k \) represents the binomial coefficient \( \binom{n}{k} \).

### Step 2: Understanding Statement 2
The second statement is:
\[ \frac{C_0}{k} - \frac{C_1}{k+1} + \frac{C_2}{k+2} - \ldots + (-1)^n \frac{C_n}{k+n} = \int_0^1 x^{k-1} (1 - x)^n \, dx \]
This is a known result related to the binomial theorem and integration.

### Step 3: Establishing the Connection
To prove Statement 1 using Statement 2, we can set \( k = 2 \) in Statement 2:
\[ \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \ldots + (-1)^n \frac{C_n}{2+n} = \int_0^1 x^{1} (1 - x)^n \, dx \]

### Step 4: Evaluating the Integral
The integral can be evaluated using the Beta function:
\[ \int_0^1 x^{1} (1 - x)^n \, dx = \frac{1 \cdot n!}{(1+n+1)!} = \frac{n!}{(n+2)!} = \frac{1}{(n+1)(n+2)} \]

### Step 5: Conclusion
Now, substituting back, we have:
\[ \frac{C_0}{2 \cdot 3} - \frac{C_1}{3 \cdot 4} + \frac{C_2}{4 \cdot 5} - \ldots + (-1)^n \frac{C_n}{(n+2)(n+3)} = \frac{1}{(n+1)(n+2)} \]
This confirms that Statement 1 is true.