Question

Let for some \(\alpha \in \mathbb{R}\), \(f: \mathbb{R} \rightarrow \mathbb{R}\) 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 \(\sum_{n=1}^{5} (\alpha + f(n))\) is:

• A. 110
• B. 140
• C. 150
• D. 170

Answer 1

The Correct Option is C

Step 1: Find the function $f(x)$
Given, \[ f(x+y)=f(x)+2y^2+y+\alpha xy \] Put $x=0$: \[ f(0+y)=f(0)+2y^2+y+\alpha(0)(y) \] \[ f(y)=f(0)+2y^2+y \] Since $f(0)=-1$, \[ f(y)=2y^2+y-1 \] Replacing $y$ by $x$, \[ f(x)=2x^2+x-1 \] Step 2: Find the value of $\alpha$
Now, \[ f(x+y)=2(x+y)^2+(x+y)-1 \] Expand: \[ f(x+y)=2(x^2+2xy+y^2)+x+y-1 \] \[ f(x+y)=2x^2+4xy+2y^2+x+y-1 \] Also from the given equation, \[ f(x)+2y^2+y+\alpha xy \] Substitute $f(x)=2x^2+x-1$: \[ = (2x^2+x-1)+2y^2+y+\alpha xy \] \[ =2x^2+x-1+2y^2+y+\alpha xy \] Comparing both expressions of $f(x+y)$, \[ 4xy=\alpha xy \] Hence, \[ \alpha=4 \] Step 3: Verify using $f(1)=2$
\[ f(1)=2(1)^2+1-1 \] \[ =2+1-1=2 \] Condition verified. Step 4: Calculate the required sum
\[ \sum_{n=1}^{5} (\alpha+f(n)) \] Substitute $\alpha=4$ and $f(n)=2n^2+n-1$: \[ = \sum_{n=1}^{5} \left(4+2n^2+n-1\right) \] \[ = \sum_{n=1}^{5} (2n^2+n+3) \] \[ =2\sum_{n=1}^{5}n^2+\sum_{n=1}^{5}n+\sum_{n=1}^{5}3 \] Use formulas: \[ \sum_{n=1}^{5} n^2=\frac{5(6)(11)}{6}=55 \] \[ \sum_{n=1}^{5} n=\frac{5(6)}{2}=15 \] \[ \sum_{n=1}^{5} 3=3\times 5=15 \] So, \[ =2(55)+15+15 \] \[ =110+30 \] \[ =140 \] Final Answer: \[ \boxed{140} \] \[ \boxed{\text{Option (B)}} \]