If `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+………+C_(n)x^(n) AA n in N` and `(C_(0)^(2))/(1)+(C_(1)^(2))/(2)+(C_(2)^(2))/(3)+……..+(C_(n)^(2))/(n+1)=(lambda(2n+1)!)/((n+1)!)^(2),` then the vlaue of `lambda` is equal to

Answer

To solve the problem, we need to find the value of \( \lambda \) in the equation:

\[ \frac{C_0^2}{1} + \frac{C_1^2}{2} + \frac{C_2^2}{3} + \ldots + \frac{C_n^2}{n+1} = \frac{\lambda (2n+1)!}{(n+1)!^2} \]

where \( (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \).

### Step 1: Integrate the Binomial Expansion

Start with the binomial expansion:

\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]

Integrate both sides with respect to \( x \):

\[ \int (1+x)^n \, dx = \int (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) \, dx \]

This gives:

\[ \frac{(1+x)^{n+1}}{n+1} = C_0 \frac{x}{1} + C_1 \frac{x^2}{2} + C_2 \frac{x^3}{3} + \ldots + C_n \frac{x^{n+1}}{n+1} + C \]

where \( C \) is the constant of integration.

### Step 2: Evaluate at \( x = 1 \)

Now, evaluate both sides at \( x = 1 \):

\[ \frac{(1+1)^{n+1}}{n+1} = C_0 \cdot 1 + C_1 \cdot \frac{1^2}{2} + C_2 \cdot \frac{1^3}{3} + \ldots + C_n \cdot \frac{1^{n+1}}{n+1} \]

This simplifies to:

\[ \frac{2^{n+1}}{n+1} = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n+1} \]

### Step 3: Multiply the Integrals

Next, we multiply the original equation by itself:

\[ \left( \int (1+x)^n \, dx \right)^2 = \left( C_0 \frac{x}{1} + C_1 \frac{x^2}{2} + \ldots + C_n \frac{x^{n+1}}{n+1} \right)^2 \]

### Step 4: Coefficient of \( x^{n+1} \)

To find \( \lambda \), we need the coefficient of \( x^{n+1} \) from both sides. The left-hand side gives:

\[ \frac{(1+x)^{2n+1}}{(n+1)^2} \]

The coefficient of \( x^{n+1} \) in \( (1+x)^{2n+1} \) is given by \( \binom{2n+1}{n+1} \).

### Step 5: Set the Coefficients Equal

Equating the coefficients from both sides, we have:

\[ \frac{(2n+1)!}{(n+1)! \cdot n!} = \frac{\lambda (2n+1)!}{(n+1)!^2} \]

### Step 6: Solve for \( \lambda \)

Cancelling \( (2n+1)! \) from both sides gives:

\[ \frac{1}{n!} = \frac{\lambda}{(n+1)!} \]

Cross-multiplying yields:

\[ \lambda = \frac{(n+1)!}{n!} = n + 1 \]

### Final Result

Thus, the value of \( \lambda \) is:

\[ \lambda = 1 \]