Evaluate Sum of Function Values Satisfying Functional Equation

To find the value of $\sum_{n=1}^{5} (\alpha + f(n))$, we first determine the explicit form of the function $f(x)$ and the value of the constant $\alpha$.

The given functional equation is: $f(x + y) = f(x) + 2y^2 + y + \alpha xy \quad \forall x, y \in \mathbb{R}$

Substitute $x = 0$ into the equation: $f(y) = f(0) + 2y^2 + y + \alpha(0)y$

Given $f(0) = -1$, we get: $f(y) = 2y^2 + y - 1$

Thus, for any $x \in \mathbb{R}$, the function is: $f(x) = 2x^2 + x - 1$

To find $\alpha$, we substitute $f(x)$ and $f(x+y)$ back into the functional equation: $2(x + y)^2 + (x + y) - 1 = (2x^2 + x - 1) + 2y^2 + y + \alpha xy$

Expanding the left side: $2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + \alpha xy$

Comparing the coefficient of $xy$ on both sides: $\alpha = 4$

We can verify $f(1) = 2(1)^2 + 1 - 1 = 2$, which satisfies the given condition $f(1) = 2$.

Now, we calculate the term inside the summation: $\alpha + f(n) = 4 + (2n^2 + n - 1) = 2n^2 + n + 3$

We need to evaluate the sum: $\sum_{n=1}^{5} (\alpha + f(n)) = \sum_{n=1}^{5} (2n^2 + n + 3)$

Using standard summation formulas: $\sum_{n=1}^{5} (\alpha + f(n)) = 2 \sum_{n=1}^{5} n^2 + \sum_{n=1}^{5} n + \sum_{n=1}^{5} 3$

Calculating each sum individually: $\sum_{n=1}^{5} n^2 = \frac{5 \times (5 + 1) \times (2 \cdot 5 + 1)}{6} = \frac{5 \times 6 \times 11}{6} = 55$

$\sum_{n=1}^{5} n = \frac{5 \times (5 + 1)}{2} = 15$

$\sum_{n=1}^{5} 3 = 3 \times 5 = 15$

Substituting these values back into the expression: $\sum_{n=1}^{5} (\alpha + f(n)) = 2(55) + 15 + 15 = 110 + 30 = 140$

Thus, the correct answer corresponds to Option B.