Let for some $$\alpha \in \mathbb{R}$$ $$f : \mathbb{R} \to \mathbb{R}$$ be a function  satisfying  $$f(x+y) = f(x) + 2y^2 + y + \alpha xy$$ for all $$x, y \in \mathbb{R}$$. If $$f(0) = -1$$ and $$f(1) = 2$$, then the value of $$\displaystyle\sum_{n=1}^{5}(\alpha + f(n))$$ is  :

We are given the functional equation
$$f(x+y)=f(x)+2y^{2}+y+\alpha xy\qquad \forall \,x,y\in\mathbb{R}$$
together with the initial values $$f(0)=-1,\; f(1)=2$$.

Step 1 : Put $$x=0$$
With $$x=0$$ the term $$\alpha xy$$ vanishes, so
$$f(y)=f(0)+2y^{2}+y = -1+2y^{2}+y.\qquad -(1)$$

Step 2 : Verify $$f(1)=2$$ and determine $$\alpha$$
From $$(1)$$, $$f(1)=-1+2(1)^{2}+1=2,$$ which agrees with the given value; hence $$(1)$$ is consistent.
Next, place $$f(t)=-1+2t^{2}+t$$ back into the original equation and compare coefficients.

Left-hand side:
$$f(x+y) = -1 + 2(x+y)^{2} + (x+y) = -1 + 2x^{2}+4xy+2y^{2}+x+y.$$

Right-hand side:
$$f(x)+2y^{2}+y+\alpha xy = \bigl(-1+2x^{2}+x\bigr) + 2y^{2}+y + \alpha xy \\[4pt] = -1 + 2x^{2}+x + 2y^{2}+y + \alpha xy.$$

Compare the $$xy$$ coefficients:
Left side gives $$4xy$$, right side gives $$\alpha xy$$. Therefore $$\alpha = 4.$$

Step 3 : Summation required
For integer $$n$$, equation $$(1)$$ gives
$$f(n)=2n^{2}+n-1.$$

We need $$\displaystyle\sum_{n=1}^{5}(\alpha+f(n)) = \sum_{n=1}^{5}\alpha + \sum_{n=1}^{5}f(n).$$

• Since $$\alpha=4$$, $$\sum_{n=1}^{5}\alpha = 5\alpha = 5\times4 = 20.$

• Compute $$$$\sum_{n=1}^{5}f(n$$)$$:
$$$$\sum_{n=1}^{5}$$ \bigl(2n^{2}+n-1\bigr) = 2$$\sum_{n=1}^{5}$$ n^{2} + $$\sum_{n=1}^{5}$$ n - $$\sum_{n=1}^{5}$$ 1.$$
Recall $$$$\sum_{n=1}^{5}$$ n^{2}=55,\; $$\sum_{n=1}^{5}$$ n = 15.$$ Hence
$$2(55)+15-5 = 110+15-5 = 120.$$

Total:
$$20 + 120 = 140.$$

Therefore the required value is $$140$$.
Option B which is: $$140$$