Quadratic Sequences

A quadratic sequence is a type of number sequence in which the difference between consecutive terms is an arithmetic sequence. In other words, the second difference (the difference between consecutive first differences) is constant.

Quadratic sequences are different from other sequences, such as arithmetic or geometric, due to their unique pattern of differences between terms.

To identify a quadratic sequence, we look at the differences between consecutive terms. If the first differences do not form a constant, but the second differences are constant, then the sequence is quadratic.

For example, let’s look at the sequence 1, 4, 9, 16, 25. The first differences are 3, 5, 7, and 9, while the second differences are 2, 2, and 2, which are constant. This confirms that the sequence is quadratic.

Finding the Nth Term of Quadratic Sequences

The general formula for the Nth term of a quadratic sequence is an^2 + bn + c, where a, b, and c are constants, and n is the term number. To get the formula for a given sequence, follow these steps:

Step 1: Calculate the first and second differences.

Step 2: The constant second difference is twice the value of ‘a’.

Step 3: Find the values of ‘b’ and ‘c’ using the first differences and the value of ‘a’ obtained in step 2.

Step 4: Write down the formula for the Nth term.

For example, let’s derive the Nth term formula for the quadratic sequence 3, 10, 19, 30, 43:

1. Calculate the first differences (7, 9, 11, 13) and second differences (2, 2, 2).

2. The constant second difference is 2, which is twice the value of ‘a’. So, a = 1.

3. Use the first differences to find ‘b’ and ‘c’. The first difference for n = 1 is 7. Plug in the values of ‘a’ and ‘n’ into the formula:

7 = 1(1) + b(1) + c

b + c = 6

The first difference for n = 2 is 9. Plug in the values of ‘a’ and ‘n’ into the formula:

9 = 1(4) + b(2) + c

2b + c = 5

We can solve this system using either substitution or elimination. Let’s use elimination:

Subtract the first equation from the second equation:

(2b + c) - (b + c) = 5 - 6

b = -1

Now, substitute the value of ‘b’ back into the first equation:

(-1) + c = 6

c = 7

So, we have found that b = -1 and c = 7.

4. Write down the formula for the Nth term: an^2 + bn + c

= n^2 - n + 7

Let’s look at some more examples:

Examples

Example 1:

Determine if the following sequence is a quadratic sequence: 5, 12, 23, 38, 57. If so, find the Nth term formula.

1. Find the first differences: 7, 11, 15, 19 Second differences: 4, 4, 4

Since the second differences are constant, it’s a quadratic sequence. The constant second difference is 4, so ‘a’ = 4/2 = 2.

2. Now, find ‘b’ and ‘c’ using the first differences and the formula an^2 + bn + can^2 + bn + c.

For n = 1: 7 = 2(1) + b(1) + c7 = 2(1) + b(1) + c

b + c = 5b + c = 5

For n = 2: 11 = 2(4) + b(2) + c11 = 2(4) + b(2) + c

2b + c = 32b + c = 3

3. Solve the system of equations:

b + c = 5b + c = 5

2b + c = 32b + c = 3

Using elimination, subtract the first equation from the second equation: b = -2b = -2

Now, substitute the value of ‘b’ back into the first equation to find ‘c’:

c = 5 - (-2)c = 5 - (-2)

c = 7c = 7

4. The Nth term formula is 2n^2 - 2n + 72n^2 - 2n + 7.

Example 2:

Find the 10th term of the quadratic sequence with an Nth term formula of n^2 + 3n + 2.

To find the 10th term, substitute n = 10 into the formula:

10^2 + 3(10) + 2 = 100 + 30 + 2 = 13210^2 + 3(10) + 2 = 100 + 30 + 2 = 132

The 10th term of the sequence is 132.

Example 3:

A scientist is studying the growth of a certain species of plants. She notices that the number of leaves on a plant increases quadratically as the plant grows.

When the plant is 1 week old, it has 5 leaves; when it is 2 weeks old, it has 11 leaves; and when it is 3 weeks old, it has 21 leaves. How many leaves will the plant have when it is 5 weeks old?

1. First, we need to find the Nth term formula for this quadratic sequence. To do this, we can start by finding the first and second differences:

First differences: 6, 10 Second differences: 4

Since the second differences are constant, it’s a quadratic sequence. The constant second difference is 4, so ‘a’ = 4/2 = 2.

2. Now, find ‘b’ and ‘c’ using the first differences and the formula an^2 + bn + can^2 + bn + c.

For n = 1: 5 = 2(1) + b(1) + c5 = 2(1) + b(1) + c

b + c = 3b + c = 3

For n = 2: 11 = 2(4) + b(2) + c11 = 2(4) + b(2) + c

2b + c = 72b + c = 7

3. Solve the system of equations:

b + c = 3b + c = 3

2b + c = 32b + c = 3

Using elimination, subtract the first equation from the second equation:

  2b + c = 3
–   b + c = 3
– – – – – – – – –
    b       = 0

Now, substitute the value of ‘b’ back into the first equation to find ‘c’:

c = 3 - 0c = 3 - 0

c = 3c = 3

The Nth term formula is 2n^2 + 32n^2 + 3.

4. Now we can find the number of leaves when the plant is 5 weeks old by plugging n = 5 into the formula:

2(5)^2 + 32(5)^2 + 3

= 2(25) + 3= 2(25) + 3

= 50 + 3 = 53= 50 + 3 = 53

When the plant is 5 weeks old, it will have 53 leaves.

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