Let f: mathbbR → mathbbR be a function defined by f(x)=(2+3 a) x^2+ ( a+2 a-1 ) x+b, a ≠ 1 . If f(x+ y)=f(x)+f( y)+1- 2 7 x y, then the value of 28…

$\begin{aligned} & \text { Put } y=0 \\ & f(x)=f(0)+f(x)+1-0 \\ & f(0)=-1 \\ & f(0)=0+0+b \\ & \Rightarrow b=-1 \\ & f(-1+1)=f(-1)+f(1)+1+\frac{2}{7} \\ & f(0)=f(-1)+f(1)+\frac{9}{7}\end{aligned}$​ Put y=0f(x)=f(0)+f(x)+1−0f(0)=−1f(0)=0+0+b⇒b=−1f(−1+1)=f(−1)+f(1)+1+72​f(0)=f(−1)+f(1)+79​​ $\begin{aligned}-1=(2+3 a)+\left(\frac{a+2}{a-1}\right)(-1)+b+(2+3 a) & \\ & +\frac{a+2}{a-1}+b+\frac{9}{7}\end{aligned}$−1=(2+3a)+(a−1a+2​)(−1)+b+(2+3a)​+a−1a+2​+b+79​​ $\displaystyle \begin{aligned} & -1=4+6 a-2+\frac{9}{7} \\ & -1=2+\frac{9}{7}+6 a \\ & 6 a=-1-2-\frac{9}{7} \\ & a=\frac{-5}{7} \\ & f(x)=\frac{-x^2}{7}+\frac{\frac{9}{7}}{\frac{-12}{7}} x-1 \\ & f(x)=\frac{-x^2}{7}-\frac{3}{4} x-1 \\ & \sum_{i=1}^5 f(i)=-\frac{1}{7}\left(\frac{5 \times 6 \times 11}{6}\right)-\frac{3}{4}\left(\frac{5 \times 6}{2}\right)-5 \\ & =\frac{-55}{7}-\frac{45}{4}-5 \\ & =\frac{675}{28} \\ & \Rightarrow 28\left|\sum_{i-1}^5 f(i)\right|=675\end{aligned}$​−1=4+6a−2+79​−1=2+79​+6a6a=−1−2−79​a=7−5​f(x)=7−x2​+7−12​79​​x−1f(x)=7−x2​−43​x−1i=1∑5​f(i)=−71​(65×6×11​)−43​(25×6​)−5=7−55​−445​−5=28675​⇒28​i−1∑5​f(i)​=675​