Let `P_n(x) =1+2x+3x^2+............+(n+1)x^n` be a polynomial such that n is even.The number of real roots of `P_n(x)=0` is

Options

A. 0
B. n
C. 1
D. n+1

Answer

To solve the problem, we need to analyze the polynomial \( P_n(x) = 1 + 2x + 3x^2 + \ldots + (n+1)x^n \) where \( n \) is even, and determine the number of real roots of the equation \( P_n(x) = 0 \).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
The polynomial \( P_n(x) \) is given by:
\[ P_n(x) = 1 + 2x + 3x^2 + \ldots + (n+1)x^n \]
Here, \( n \) is even, which means \( n = 2k \) for some integer \( k \).

2. **Evaluating at Specific Points**:
Let's evaluate \( P_n(x) \) at \( x = 0 \):
\[ P_n(0) = 1 + 2(0) + 3(0)^2 + \ldots + (n+1)(0)^n = 1 \]
This shows that \( P_n(0) = 1 \), which is positive.

3. **Behavior as \( x \to \infty \)**:
As \( x \) approaches infinity, the term \( (n+1)x^n \) dominates the polynomial since it has the highest degree. Therefore:
\[ P_n(x) \to \infty \quad \text{as} \quad x \to \infty \]

4. **Behavior as \( x \to -\infty \)**:
As \( x \) approaches negative infinity, since \( n \) is even, the leading term \( (n+1)x^n \) will also approach infinity:
\[ P_n(x) \to \infty \quad \text{as} \quad x \to -\infty \]

5. **Finding Critical Points**:
To find the critical points, we can compute the derivative \( P_n'(x) \):
\[ P_n'(x) = 2 + 3 \cdot 2x + 4 \cdot 3x^2 + \ldots + (n+1) \cdot n x^{n-1} \]
Since all coefficients are positive, \( P_n'(x) > 0 \) for all \( x \). This indicates that \( P_n(x) \) is a strictly increasing function.

6. **Conclusion on Roots**:
Since \( P_n(x) \) is strictly increasing, and we have established that \( P_n(0) = 1 > 0 \) and \( P_n(x) \to \infty \) as \( x \to \infty \) and \( x \to -\infty \), it follows that \( P_n(x) \) does not cross the x-axis. Therefore, the polynomial \( P_n(x) = 0 \) has no real roots.

### Final Answer:
The number of real roots of \( P_n(x) = 0 \) is **0**.