If `n_1, n_2, n_3 ,……., n_k` are such that, out of these k elements `k/2` elements are even and rest are odd numbers. Which is necessarily even?

Options

A. `(n_1 + 2n_2 + 3n_3 + 4 n_4 + …..+ k.n_k)`
B. `n_1.n_2.n_3……n_((k)/2 + 1)`
C. `(n_1 + n_3 + n_5 + n_7 + ….+ n_(k-1))`
D. `n_2 + n_4 + n_6 + ……. + n_k`

Answer

To solve the problem, we need to analyze the given information about the numbers and determine which option is necessarily even.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We have `k` elements where `k/2` are even and `k/2` are odd. We need to find which of the given options is necessarily even.

2. **Identifying the Elements:**
Let's denote the elements as `n1, n2, n3, ..., nk`. Since half of them are even and half are odd, we can represent them as follows:
- Even numbers: `e1, e2, ..., e(k/2)`
- Odd numbers: `o1, o2, ..., o(k/2)`

3. **Analyzing the Options:**
We need to evaluate each option to see if it results in an even number regardless of the specific values of `n1, n2, ..., nk`.

4. **Case Analysis:**
- **Case 1:** Let's take an example where `k = 6`. So we have 3 even numbers and 3 odd numbers.
- Example: `1, 2, 3, 4, 5, 6` (Odds: 1, 3, 5; Evens: 2, 4, 6)
- **Case 2:** Another example with `k = 6` could be `2, 3, 4, 5, 6, 7` (Odds: 3, 5, 7; Evens: 2, 4, 6)

5. **Evaluating Each Option:**
- For each option provided, we need to check if the result is even for both cases.

6. **Finding the Necessary Even Result:**
- After evaluating the options, we find that one option consistently results in an even number regardless of the arrangement of odd and even numbers.

7. **Conclusion:**
- The option that is necessarily even is the second option, as it produces an even result in both cases analyzed.

### Final Answer:
The answer is **Option 2**.