Key Concepts and Formulas

• Functional Equations: Properties of functions derived from equations involving $f(x+y)$, $f(x)$, $f(y)$, etc. Specific substitutions like $x=0, y=0$ or $y=-x$ are crucial.
• Polynomial Identity: If a polynomial is identically zero for all real values of the variable, then all its coefficients must be zero.
• Quadratic Functions: The form $f(x) = Ax^2 + Bx + C$, and how to find its coefficients.
• Summation Formulas: Basic summation properties and the ability to evaluate sums of terms.

Step-by-Step Solution

Step 1: Determine the value of $f(0)$ using the functional equation.

We are given the functional equation: $f(x+y) = f(x) + f(y) + 1 - \frac{2}{7} xy \quad \text{.... (1)}$ To find $f(0)$, we substitute $x=0$ and $y=0$ into equation (1). $f(0+0) = f(0) + f(0) + 1 - \frac{2}{7}(0)(0)$ $f(0) = 2f(0) + 1$ Subtracting $f(0)$ from both sides gives: $0 = f(0) + 1$ Therefore, $f(0) = -1$

Step 2: Determine the value of the constant $b$ from the explicit form of $f(x)$.

The function is given by $f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+b$. We can find $f(0)$ by substituting $x=0$ into this explicit form: $f(0) = (2+3a)(0)^2 + \left(\frac{a+2}{a-1}\right)(0) + b$ $f(0) = 0 + 0 + b$ $f(0) = b$ From Step 1, we know $f(0) = -1$. Equating the two expressions for $f(0)$, we get: $b = -1$

Step 3: Determine the value of the constant $a$ using the functional equation and the form of $f(x)$.

We will now use the substitution $y=-x$ in the functional equation (1). $f(x+(-x)) = f(x) + f(-x) + 1 - \frac{2}{7} x(-x)$ $f(0) = f(x) + f(-x) + 1 + \frac{2}{7} x^2$ Since $f(0) = -1$, we have: $-1 = f(x) + f(-x) + 1 + \frac{2}{7} x^2$ Rearranging this equation, we get: $f(x) + f(-x) = -2 - \frac{2}{7} x^2 \quad \text{.... (2)}$

Now, let's use the explicit form of $f(x)$ with $b=-1$: $f(x) = (2+3a)x^2 + \left(\frac{a+2}{a-1}\right)x - 1$ Let $A = 2+3a$ and $B = \frac{a+2}{a-1}$. So, $f(x) = Ax^2 + Bx - 1$. Then, $f(-x)$ is: $f(-x) = A(-x)^2 + B(-x) - 1 = Ax^2 - Bx - 1$ Now, we compute $f(x) + f(-x)$: $f(x) + f(-x) = (Ax^2 + Bx - 1) + (Ax^2 - Bx - 1)$ $f(x) + f(-x) = 2Ax^2 - 2$ Substitute $A = 2+3a$: $f(x) + f(-x) = 2(2+3a)x^2 - 2$ Now, substitute this expression for $f(x) + f(-x)$ into equation (2): $2(2+3a)x^2 - 2 = -2 - \frac{2}{7} x^2$ Adding 2 to both sides, we get: $2(2+3a)x^2 = -\frac{2}{7} x^2$ This equation must hold for all real values of $x$. Therefore, the coefficients of $x^2$ on both sides must be equal. $2(2+3a) = -\frac{2}{7}$ Divide both sides by 2: $2+3a = -\frac{1}{7}$ $3a = -\frac{1}{7} - 2$ $3a = -\frac{1}{7} - \frac{14}{7}$ $3a = -\frac{15}{7}$ $a = -\frac{15}{7} \cdot \frac{1}{3}$ $a = -\frac{5}{7}$

Step 4: Reconstruct the function $f(x)$ with the determined values of $a$ and $b$.

We have $a = -\frac{5}{7}$ and $b = -1$. Let's find the coefficients of $f(x)$:

• Coefficient of $x^2$: $2+3a = 2 + 3\left(-\frac{5}{7}\right) = 2 - \frac{15}{7} = \frac{14-15}{7} = -\frac{1}{7}$.
• Coefficient of $x$: $\frac{a+2}{a-1} = \frac{-\frac{5}{7}+2}{-\frac{5}{7}-1} = \frac{\frac{-5+14}{7}}{\frac{-5-7}{7}} = \frac{\frac{9}{7}}{\frac{-12}{7}} = \frac{9}{-12} = -\frac{3}{4}$.
• Constant term: $b = -1$.

So, the function $f(x)$ is: $f(x) = -\frac{1}{7} x^2 - \frac{3}{4} x - 1$ To simplify calculations for the sum, we find a common denominator for the coefficients, which is 28. $f(x) = -\frac{4}{28} x^2 - \frac{21}{28} x - \frac{28}{28}$ $f(x) = -\frac{1}{28} (4x^2 + 21x + 28)$

Step 5: Calculate the values of $|f(i)|$ for $i=1, 2, 3, 4, 5$.

We need to calculate $|f(i)| = \left|-\frac{1}{28} (4i^2 + 21i + 28)\right| = \frac{1}{28} |4i^2 + 21i + 28|$. For $i \ge 1$, the expression $4i^2 + 21i + 28$ is always positive because all terms are positive. Thus, $|4i^2 + 21i + 28| = 4i^2 + 21i + 28$. So, $|f(i)| = \frac{1}{28} (4i^2 + 21i + 28)$.

Let's calculate for each $i$:

• For $i=1$: $|f(1)| = \frac{1}{28} (4(1)^2 + 21(1) + 28) = \frac{1}{28} (4 + 21 + 28) = \frac{53}{28}$.
• For $i=2$: $|f(2)| = \frac{1}{28} (4(2)^2 + 21(2) + 28) = \frac{1}{28} (4(4) + 42 + 28) = \frac{1}{28} (16 + 42 + 28) = \frac{86}{28}$.
• For $i=3$: $|f(3)| = \frac{1}{28} (4(3)^2 + 21(3) + 28) = \frac{1}{28} (4(9) + 63 + 28) = \frac{1}{28} (36 + 63 + 28) = \frac{127}{28}$.
• For $i=4$: $|f(4)| = \frac{1}{28} (4(4)^2 + 21(4) + 28) = \frac{1}{28} (4(16) + 84 + 28) = \frac{1}{28} (64 + 84 + 28) = \frac{176}{28}$.
• For $i=5$: $|f(5)| = \frac{1}{28} (4(5)^2 + 21(5) + 28) = \frac{1}{28} (4(25) + 105 + 28) = \frac{1}{28} (100 + 105 + 28) = \frac{233}{28}$.

Step 6: Calculate the sum $\sum\limits_{i=1}^5|f(i)|$.

$\sum\limits_{i=1}^5|f(i)| = |f(1)| + |f(2)| + |f(3)| + |f(4)| + |f(5)|$ $\sum\limits_{i=1}^5|f(i)| = \frac{53}{28} + \frac{86}{28} + \frac{127}{28} + \frac{176}{28} + \frac{233}{28}$ $\sum\limits_{i=1}^5|f(i)| = \frac{53 + 86 + 127 + 176 + 233}{28}$ $\sum\limits_{i=1}^5|f(i)| = \frac{675}{28}$

Step 7: Calculate the final value $28 \sum\limits_{i=1}^5|f(i)|$.

$28 \sum\limits_{i=1}^5|f(i)| = 28 \times \frac{675}{28}$ $28 \sum\limits_{i=1}^5|f(i)| = 675$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and signs.
• Sign of the Quadratic: Always check if the quadratic expression inside the absolute value is positive or negative for the given range of $x$. In this case, $4i^2 + 21i + 28$ is positive for $i \ge 1$.
• Substitution Strategy: The choice of substitutions in functional equations is key. $x=0, y=0$ and $y=-x$ are standard and effective for quadratic forms.

Summary

The problem involves a functional equation and an explicit quadratic form of a function. We first used substitutions in the functional equation to find $f(0)$ and then related it to the constant term $b$. By substituting $y=-x$ and comparing coefficients of the resulting polynomial identity, we determined the value of $a$. With $a$ and $b$ found, we reconstructed the function $f(x)$ and then computed the sum of $|f(i)|$ for $i=1$ to $5$, finally multiplying by 28 to get the desired result.

The final answer is $\boxed{675}$.