`C_(0)^(2)+1/2C_(1)^(2)+1/3C_(2)^(2)+….+1/(n+1)C_(n)^(2)` equals

Options

A. `1/n(""^(2n)C_(n))`
B. `1/(n+1)(""^(2n+1)C_(n))`
C. `1/n(""^(2n)C_(n+1))`
D. `2^(n)`

Answer

To solve the problem \( C_0^2 + \frac{1}{2} C_1^2 + \frac{1}{3} C_2^2 + \ldots + \frac{1}{n+1} C_n^2 \), we will use properties of binomial coefficients and summation techniques. Here’s the step-by-step solution:

### Step 1: Understand the terms involved
The term \( C_r^2 \) refers to the binomial coefficient, which can be expressed as \( C_r^2 = \binom{n}{r}^2 \).

### Step 2: Rewrite the expression
We can rewrite the given expression as:
\[ \sum_{r=0}^{n} \frac{1}{r+1} C_r^2 \]
where \( C_r = \binom{n}{r} \).

### Step 3: Use the identity for binomial coefficients
There is a known identity that states:
\[ \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \]
This identity will help us relate the squared binomial coefficients to a single binomial coefficient.

### Step 4: Apply the identity to our sum
Using the identity, we can express our sum as:
\[ \sum_{r=0}^{n} \frac{1}{r+1} C_r^2 = \frac{1}{n+1} \sum_{r=0}^{n} C_r^2 \]
Thus, we have:
\[ \sum_{r=0}^{n} \frac{1}{r+1} C_r^2 = \frac{1}{n+1} \binom{2n}{n} \]

### Step 5: Final expression
Therefore, the final expression for the original sum is:
\[ C_0^2 + \frac{1}{2} C_1^2 + \frac{1}{3} C_2^2 + \ldots + \frac{1}{n+1} C_n^2 = \frac{1}{n+1} \binom{2n}{n} \]

### Conclusion
The result is:
\[ \frac{1}{n+1} \binom{2n}{n} \]

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