Question1. Let $a_{n}=n^{2}-16 n+10$ where $n \in \mathbb{N}$ . What is the smallest $n_{0} \in \mathbb{N}$ such that $a_{n}$ is increasing for all $n \geqslant n_{0}$ ?

Studdy Solution

STEP 1

1. The sequence is given by $a_n = n^2 - 16n + 10$ .
2. We need to find when the sequence $a_n$ starts increasing.
3. The sequence is increasing when the difference between consecutive terms is positive.

STEP 2

1. Express the condition for the sequence to be increasing.
2. Find the difference between consecutive terms.
3. Solve the inequality to find the smallest $n_0$ .

STEP 3

To determine when the sequence $a_n$ is increasing, we need to find when:
$a_{n+1} - a_n > 0$

STEP 4

Calculate $a_{n+1}$ and then find the difference $a_{n+1} - a_n$ :
$a_{n+1} = (n+1)^2 - 16(n+1) + 10$
$a_{n+1} = (n^2 + 2n + 1) - 16n - 16 + 10$
$a_{n+1} = n^2 + 2n + 1 - 16n - 16 + 10$
$a_{n+1} = n^2 - 14n - 5$
Now, calculate the difference:
$a_{n+1} - a_n = (n^2 - 14n - 5) - (n^2 - 16n + 10)$
$a_{n+1} - a_n = n^2 - 14n - 5 - n^2 + 16n - 10$
$a_{n+1} - a_n = 2n - 15$

STEP 5

Solve the inequality $2n - 15 > 0$ to find the smallest $n_0$ :
$2n - 15 > 0$
$2n > 15$
$n > \frac{15}{2}$
Since $n$ is a natural number, the smallest integer greater than $\frac{15}{2}$ is $n = 8$ .
Thus, the smallest $n_0$ is $\boxed{8}$ .

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