If `sum_(i=1)^(n)a_(i)^(2)=lambda, AAa_(i)ge0` and if greatest and least values of `(sum_(i=1)^(n)a_(i))^(2)` are `lambda_(1)` and `lambda_(2)` respectively, then `(lambda_(1)-lambda_(2))` is

Options

A. `nlambda`
B. `(n-1)lambda`
C. `(n+2)lambda`
D. `(n+1)lambda`

Explanation

To solve the problem, we need to find the difference between the greatest and least values of \((\sum_{i=1}^{n} a_i)^2\) given that \(\sum_{i=1}^{n} a_i^2 = \lambda\) and \(a_i \geq 0\).

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that \(\sum_{i=1}^{n} a_i^2 = \lambda\). We want to find the maximum and minimum values of \((\sum_{i=1}^{n} a_i)^2\).

2. **Applying Cauchy-Schwarz Inequality**:
By the Cauchy-Schwarz inequality, we have:
\[ \left( \sum_{i=1}^{n} a_i \right)^2 \leq n \sum_{i=1}^{n} a_i^2 \]
Substituting \(\sum_{i=1}^{n} a_i^2 = \lambda\), we get:
\[ \left( \sum_{i=1}^{n} a_i \right)^2 \leq n\lambda \]
This gives us the upper bound for \((\sum_{i=1}^{n} a_i)^2\).

3. **Finding the Maximum Value**:
The maximum value of \((\sum_{i=1}^{n} a_i)^2\) occurs when all \(a_i\) are equal. If \(a_1 = a_2 = \ldots = a_n = k\), then:
\[ n k^2 = \lambda \implies k^2 = \frac{\lambda}{n} \implies k = \sqrt{\frac{\lambda}{n}} \]
Therefore,
\[ \sum_{i=1}^{n} a_i = n \cdot \sqrt{\frac{\lambda}{n}} = \sqrt{n\lambda} \]
Thus, the maximum value is:
\[ \lambda_1 = n\lambda \]

4. **Finding the Minimum Value**:
The minimum value of \((\sum_{i=1}^{n} a_i)^2\) occurs when one \(a_i\) is \(\sqrt{\lambda}\) and the rest are 0. Thus:
\[ \sum_{i=1}^{n} a_i = \sqrt{\lambda} \]
Therefore, the minimum value is:
\[ \lambda_2 = \lambda \]

5. **Calculating the Difference**:
Now, we can find the difference:
\[ \lambda_1 - \lambda_2 = n\lambda - \lambda = (n - 1)\lambda \]

### Final Answer:
\[ \lambda_1 - \lambda_2 = (n - 1)\lambda \]