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 start with the given expression:

\[ (1+x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + \ldots + C_{n}x^{n} \]

where \( C_{r} = \binom{n}{r} \) for \( r = 0, 1, 2, \ldots, n \).

We need to evaluate the sum:

\[ \frac{C_{0}^{2}}{1} + \frac{C_{1}^{2}}{2} + \frac{C_{2}^{2}}{3} + \ldots + \frac{C_{n}^{2}}{n+1} \]

This can be expressed as:

\[ \sum_{r=0}^{n} \frac{C_{r}^{2}}{r+1} \]

Using the identity \( C_{r}^{2} = \binom{n}{r} \binom{n}{r} \), we can rewrite the sum as:

\[ \sum_{r=0}^{n} \frac{\binom{n}{r} \binom{n}{r}}{r+1} \]

Next, we can use the identity:

\[ \frac{C_{r}^{2}}{r+1} = \frac{1}{n+1} \binom{n+1}{r+1} \binom{n}{r} \]

This allows us to express the sum as:

\[ \frac{1}{n+1} \sum_{r=0}^{n} \binom{n+1}{r+1} \binom{n}{r} \]

Now, we can change the index of summation by letting \( k = r + 1 \), which gives us:

\[ \frac{1}{n+1} \sum_{k=1}^{n+1} \binom{n+1}{k} \binom{n}{k-1} \]

Using the Vandermonde identity, we know that:

\[ \sum_{k=0}^{m} \binom{x}{k} \binom{y}{m-k} = \binom{x+y}{m} \]

Applying this identity here, we have:

\[ \sum_{k=1}^{n+1} \binom{n+1}{k} \binom{n}{k-1} = \binom{2n+1}{n} \]

Thus, we can rewrite our sum as:

\[ \frac{1}{n+1} \binom{2n+1}{n} \]

Now, substituting this back into the original equation, we have:

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

Next, we simplify \( \binom{2n+1}{n} \):

\[ \binom{2n+1}{n} = \frac{(2n+1)!}{n!(n+1)!} \]

Substituting this into our equation gives:

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

Cancelling \( (2n+1)! \) from both sides, we get:

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

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

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

### Summary of Steps:
1. Start with the binomial expansion of \( (1+x)^{n} \).
2. Write the sum \( \sum_{r=0}^{n} \frac{C_{r}^{2}}{r+1} \).
3. Use the identity for \( C_{r}^{2} \) to rewrite the sum.
4. Change the index of summation and apply the Vandermonde identity.
5. Simplify to find \( \lambda \).