The following sequence of numbers forms a quadratic sequence: $$ -3 ;-2 ;-3 ;-6 ;-11 ; \ldots $$ 2.2.1 The FIRST differences of the above sequence also form a sequence. Determine an expression for the general term of the first differences. 2.2 .2 Calculate the first difference between the $$ 35^{ ext {th }} $$ and $$ 36^{ ext {th }} $$ terms of the quadratic sequence. 2.2.3 Determine an expression for the $$ n^{ ext {th }} $$ term of the quadratic sequence. 2.2.4 Show that the sequence of numbers will NEVER contain a positive term.

Question

The following sequence of numbers forms a quadratic sequence: $$ -3 ;-2 ;-3 ;-6 ;-11 ; \ldots $$ 2.2.1 The FIRST differences of the above sequence also form a sequence. Determine an expression for the general term of the first differences. 2.2 .2 Calculate the first difference between the $$ 35^{	ext {th }} $$ and $$ 36^{	ext {th }} $$ terms of the quadratic sequence. 2.2.3 Determine an expression for the $$ n^{	ext {th }} $$ term of the quadratic sequence. 2.2.4 Show that the sequence of numbers will NEVER contain a positive term.

UpStudy AI · Accepted Answer

To solve the problem step by step, let's start with the given quadratic sequence:

$$
-3, -2, -3, -6, -11, \ldots
$$

### Step 1: Calculate the First Differences

The first differences of a sequence are found by subtracting each term from the next term. Let's calculate the first differences for the given sequence:

1. $$ -2 - (-3) = 1 $$
2. $$ -3 - (-2) = -1 $$
3. $$ -6 - (-3) = -3 $$
4. $$ -11 - (-6) = -5 $$

So, the first differences are:

$$
1, -1, -3, -5, \ldots
$$

### 2.2.1 Determine an Expression for the General Term of the First Differences

Now, let's find the pattern in the first differences:

- The first differences are: $$ 1, -1, -3, -5 $$

Calculating the second differences:

1. $$ -1 - 1 = -2 $$
2. $$ -3 - (-1) = -2 $$
3. $$ -5 - (-3) = -2 $$

The second differences are constant and equal to $$-2$$, indicating that the first differences form an arithmetic sequence.

The general term of the first differences can be expressed as:

$$
d_n = 1 - 2(n-1) = 3 - 2n
$$

### 2.2.2 Calculate the First Difference Between the $$ 35^{\text{th}} $$ and $$ 36^{\text{th}} $$ Terms

Using the expression for the first differences, we can calculate $$ d_{35} $$:

$$
d_{35} = 3 - 2 \cdot 35 = 3 - 70 = -67
$$

Thus, the first difference between the $$ 35^{\text{th}} $$ and $$ 36^{\text{th}} $$ terms is $$-67$$.

### 2.2.3 Determine an Expression for the $$ n^{\text{th}} $$ Term of the Quadratic Sequence

To find the $$ n^{\text{th}} $$ term of the quadratic sequence, we can use the first differences. The $$ n^{\text{th}} $$ term can be expressed as:

$$
a_n = a_1 + \sum_{k=1}^{n-1} d_k
$$

Where $$ a_1 = -3 $$ and $$ d_k = 3 - 2k $$.

Calculating the sum of the first differences:

$$
\sum_{k=1}^{n-1} d_k = \sum_{k=1}^{n-1} (3 - 2k) = 3(n-1) - 2\sum_{k=1}^{n-1} k
$$

Using the formula for the sum of the first $$ m $$ integers:

$$
\sum_{k=1}^{m} k = \frac{m(m+1)}{2}
$$

We have:

$$
\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}
$$

Thus,

$$
\sum_{k=1}^{n-1} d_k = 3(n-1) - 2 \cdot \frac{(n-1)n}{2} = 3(n-1) - (n-1)n = (n-1)(3 - n)
$$

Now substituting back into the expression for $$ a_n $$:

$$
a_n = -3 + (n-1)(3 - n)
$$

Expanding this gives:

$$
a_n = -3 + (3n - n^2 - 3 + n) = -n^2 + 4n - 6
$$

Thus, the $$ n^{\text{th}} $$ term of the quadratic sequence is:

$$
a_n = -n^2 + 4n - 6
$$

### 2.2.4 Show that the Sequence of Numbers Will NEVER Contain a Positive Term

To determine if the sequence can ever be positive, we need to analyze the quadratic expression:

$$
-n^2 + 4n - 6
$$

This is a downward-opening parabola. To find the maximum value, we can use the vertex formula $$ n = -\frac{b}{2a} $$:

$$
n = -\frac{4}{2 \cdot -1} = 2
$$

Now, substituting $$ n = 2 $$ into the expression:

$$
a_2 = -2^2 + 4 \cdot 2 - 6 = -4 + 8 - 6 = -2
$$

The maximum value of the sequence occurs at $$ n = 2 $$ and is $$-2$$. Since the parabola opens downwards and the maximum value is negative, the sequence will never contain a positive term.

### Final Summary

1. The expression for the first differences is $$ d_n = 3 - 2n $$.
2. The first difference between the $$ 35^{\text{th}} $$ and $$ 36^{\text{th}} $$ terms is $$-67$$.
3. The $$ n^{\text{th}} $$ term of the quadratic sequence is $$ a_n = -n^2 + 4n - 6 $$.
4. The sequence will never contain a positive term.
The sequence's first differences are $$ d_n = 3 - 2n $$. The first difference between the 35th and 36th terms is -67. The nth term of the quadratic sequence is $$ a_n = -n^2 + 4n - 6 $$. The sequence will never have a positive term.

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