If `C_(r)` stands for `.^(n)C_(r)=(n!)/(r!n-r!) and underset(r=1)overset(n)(sum)r.C_(r)^(2)=lamda` for `n ge 2`, then `lamda` is divisible by

Options

A. `3(n-1)`
B. `n+1`
C. `n(2n-1)`
D. `n^(2)+1`

Answer

To solve the problem, we need to evaluate the expression given by the summation \( \sum_{r=1}^{n} r \cdot C(n, r)^2 \) and find out what \( \lambda \) is divisible by.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ \lambda = \sum_{r=1}^{n} r \cdot C(n, r)^2 \]
where \( C(n, r) \) is the binomial coefficient defined as:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

2. **Rewriting the Summation**:
We can rewrite \( r \cdot C(n, r) \) as:
\[ r \cdot C(n, r) = n \cdot C(n-1, r-1) \]
This substitution allows us to express the summation in terms of \( C(n-1, r-1) \):
\[ \lambda = \sum_{r=1}^{n} n \cdot C(n-1, r-1) \cdot C(n, r) \]

3. **Changing the Index of Summation**:
By changing the index of summation (let \( k = r - 1 \)), we get:
\[ \lambda = n \sum_{k=0}^{n-1} C(n-1, k) \cdot C(n-1, k) \]
This is equivalent to:
\[ \lambda = n \cdot \sum_{k=0}^{n-1} C(n-1, k)^2 \]

4. **Using the Identity for Binomial Coefficients**:
There is a known identity for the sum of squares of binomial coefficients:
\[ \sum_{k=0}^{m} C(m, k)^2 = C(2m, m) \]
Applying this identity, we have:
\[ \sum_{k=0}^{n-1} C(n-1, k)^2 = C(2(n-1), n-1) \]

5. **Final Expression for \( \lambda \)**:
Substituting back, we get:
\[ \lambda = n \cdot C(2(n-1), n-1) \]

6. **Finding Divisibility**:
Now, we need to determine what \( \lambda \) is divisible by. The term \( C(2(n-1), n-1) \) counts the number of ways to choose \( n-1 \) elements from \( 2(n-1) \) elements, and it is always an integer. The factor \( n \) indicates that \( \lambda \) is divisible by \( n \).

### Conclusion:
Thus, \( \lambda \) is divisible by \( n \).