Statement -1: ` sum_(r=0)^(n) r(""^(n)C_(r))^(2) = n (""^(2n -1)C_(n-1))`
Statement-2: ` sum_(r=0)^(n) (""^(n)C_(r))^(2)= ""^(2n)C_(n)`

Options

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

Explanation

To solve the problem, we need to verify the two statements given:

1. **Statement 1**: \(\sum_{r=0}^{n} r \binom{n}{r}^2 = n \binom{2n-1}{n-1}\)
2. **Statement 2**: \(\sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n}\)

### Step-by-Step Solution:

**Step 1: Verify Statement 1**

We start with the left-hand side of Statement 1:

\[ \sum_{r=0}^{n} r \binom{n}{r}^2 \]

Using the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \), we can rewrite the summation:

\[ \sum_{r=0}^{n} r \binom{n}{r}^2 = \sum_{r=1}^{n} n \binom{n-1}{r-1} \binom{n}{r} \]

This can be simplified further:

\[ = n \sum_{r=1}^{n} \binom{n-1}{r-1} \binom{n}{r} \]

Changing the index of summation by letting \( k = r - 1 \):

\[ = n \sum_{k=0}^{n-1} \binom{n-1}{k} \binom{n}{k+1} \]

Using Vandermonde's identity, we know that:

\[ \sum_{k=0}^{m} \binom{a}{k} \binom{b}{m-k} = \binom{a+b}{m} \]

Applying this identity here gives:

\[ = n \binom{2n-1}{n-1} \]

Thus, we have verified Statement 1:

\[ \sum_{r=0}^{n} r \binom{n}{r}^2 = n \binom{2n-1}{n-1} \]

**Step 2: Verify Statement 2**

Now we check Statement 2:

\[ \sum_{r=0}^{n} \binom{n}{r}^2 \]

Using the identity that states the sum of the squares of binomial coefficients is given by:

\[ \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \]

This is a well-known result derived from the expansion of \((1+x)^n(1+x)^n = (1+x)^{2n}\). The coefficient of \(x^n\) in this expansion is \(\binom{2n}{n}\).

Thus, we have verified Statement 2:

\[ \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \]

### Conclusion

Both statements are true:

- **Statement 1** is true: \(\sum_{r=0}^{n} r \binom{n}{r}^2 = n \binom{2n-1}{n-1}\)
- **Statement 2** is true: \(\sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n}\)

However, Statement 2 does not serve as a correct explanation for Statement 1.