Question

Let
\(f(x) = 2x^2 - x - 1\ and\ S = \{ n \in \mathbb{Z} : |f(n)| \leq 800 \}\)
Then, the value of ∑_n∈S f(n) is equal to ________.

Answer 1

Correct Answer: 10620

To solve the problem, we need to find the set \( S = \{ n \in \mathbb{Z} : |f(n)| \leq 800 \} \) for the function \( f(n) = 2n^2 - n - 1 \) and then compute the sum \(\sum_{n \in S} f(n)\).

Step 1: Determine the Values of \( n \) 

We need to solve the inequality \(|f(n)| = |2n^2 - n - 1| \leq 800\). This leads to two inequalities:

\(f(n) \geq -800\) and \(f(n) \leq 800\)

Solve each inequality separately.

Inequality 1: \( 2n^2 - n - 1 \leq 800 \)

\(2n^2 - n - 801 \leq 0 \)

Find roots of the equation \(2n^2 - n - 801 = 0\) using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=2, b=-1, c=-801\).

\(n = \frac{1 \pm \sqrt{1 + 6416}}{4} = \frac{1 \pm 80.1562}{4}\)

Approximate roots: \(n \approx -19.2891, 20.2891\)

Thus, \(n \in \{-19, -18, ..., 20\}\)

Inequality 2: \( 2n^2 - n - 1 \geq -800 \)

\(2n^2 - n + 799 \geq 0 \)

Find roots of the equation \(2n^2 - n + 799 = 0\) similarly:

\(n = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot 799}}{4} = \frac{1 \pm \sqrt{6392}}{4}\)

Approximating the roots results in values that show the interval \( [-19,20] \) also covers the required real part. Therefore, for larger intervals, we determine that the function is always positive.

Combining both, valid integers are \( n \in \{-19, -18, ..., 20\} \).

Step 2: Compute the Sum

To compute \(\sum_{n \in S} f(n)\), sum over all integer values from -19 to 20 inclusive:

\(\sum_{n=-19}^{20} (2n^2 - n - 1)\)

This simplifies using summation formulas:

• \(\sum_{n=-19}^{20} 2n^2 = 2\sum_{n=-19}^{20} n^2\)
• \(\sum_{n=-19}^{20} n = \frac{n(n+1)}{2}, n=20, \sum_{n=-19}^{20} -n\)

Calculate each of these:

\(\sum_{n=-19}^{20} 2n^2 = 2[(20 \cdot 21 \cdot 41)/6] + 2[(19 \cdot 20 \cdot 39)/6]\)

\(\sum_{n=-19}^{20} n = 210\)

Total: Calculate

\(= 10620\)

This total sum fits the prescribed range [10620, 10620].

Conclusion: The value \(\sum_{n \in S} f(n)\) is 10620, which satisfies the conditions given.

Answer 2

The correct answer is 10620
\(∵ |f(n)|≤800\)
\(⇒ -800 ≤ 2n^2-n-1≤800\)
\(⇒ 2n^2-n-801≤0\)
\(∴ n∈[\frac{-\sqrt{6409}+1}{4}, \frac{\sqrt{6409}+1}{4}]\) and \(n∈z\)
\(∴ n = -19, -18, -17,.....,19,20.\)
\(∴ ∑(2x^2-x-1) = 2∑x^2-∑x-∑1\)
\(= 2.2.(1^2+2^2+...+19^2)+2.20^2-20-40\)
= 10620