If `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+…+C_(n)x^(n), Sigma_(r=0)^(n)((r+1)^(2))C_(r)=2^(n-2)f(n)` and if the roots of the equation f(x) = 0 are `alpha` and `beta`, then the value of `alpha^(2)+beta^(2)` is equal to (where `C_(r)` denotes `.^(n)C_(r)`)

Options

A. 13
B. 10
C. 17
D. 20

Answer

To solve the given problem, we will follow a systematic approach. The problem involves using properties of binomial coefficients and polynomial identities.

### Step-by-Step Solution:

1. **Understand the Given Equation**:
We start with the equation:
\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
where \( C_r = \binom{n}{r} \).

2. **Rewrite the Summation**:
We need to evaluate the summation:
\[ \sum_{r=0}^{n} (r+1)^2 C_r = 2^{n-2} f(n) \]
We can expand \( (r+1)^2 \) as:
\[ (r+1)^2 = r^2 + 2r + 1 \]
Therefore, we can rewrite the summation as:
\[ \sum_{r=0}^{n} (r^2 + 2r + 1) C_r = \sum_{r=0}^{n} r^2 C_r + 2 \sum_{r=0}^{n} r C_r + \sum_{r=0}^{n} C_r \]

3. **Evaluate Each Term**:
- The last term \( \sum_{r=0}^{n} C_r \) is simply \( 2^n \).
- The term \( \sum_{r=0}^{n} r C_r \) can be evaluated using the identity:
\[ \sum_{r=0}^{n} r C_r = n \cdot 2^{n-1} \]
- For the term \( \sum_{r=0}^{n} r^2 C_r \), we can use the identity:
\[ \sum_{r=0}^{n} r^2 C_r = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} \]

4. **Combine the Results**:
Now substituting the evaluated terms back into the summation:
\[ \sum_{r=0}^{n} (r+1)^2 C_r = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} + 2^n \]
Simplifying this gives:
\[ = n(n-1) \cdot 2^{n-2} + 2n \cdot 2^{n-2} + 2^n \]
\[ = (n^2 - n + 2n) \cdot 2^{n-2} + 2^n = n^2 + n \cdot 2^{n-2} + 2^n \]

5. **Set Equal to Given Expression**:
From the problem statement, we have:
\[ n^2 + n \cdot 2^{n-2} + 2^n = 2^{n-2} f(n) \]
This implies:
\[ f(n) = n^2 + 3n + 4 \]

6. **Find Roots of the Equation**:
We need to find the roots of the equation \( f(x) = 0 \):
\[ x^2 + 3x + 4 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 16}}{2} = \frac{-3 \pm \sqrt{-7}}{2} \]
The roots are:
\[ \alpha = \frac{-3 + i\sqrt{7}}{2}, \quad \beta = \frac{-3 - i\sqrt{7}}{2} \]

7. **Calculate \( \alpha^2 + \beta^2 \)**:
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
- \( \alpha + \beta = -3 \)
- \( \alpha \beta = 4 \)

Thus:
\[ \alpha^2 + \beta^2 = (-3)^2 - 2 \cdot 4 = 9 - 8 = 1 \]

### Final Result:
The value of \( \alpha^2 + \beta^2 \) is \( 1 \).