Math Problem Statement
Solution
Let's work through this sequence step by step!
1. First and Second Differences:
The first difference is found by subtracting consecutive terms in the original sequence.
- First difference: , , , , and .
The second difference is found by subtracting consecutive terms of the first difference.
- Second difference: Similar steps as above, using the values obtained from the first difference.
Let’s calculate:
Original Sequence: 11, 25, 45, 71, 103, 141
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First difference:
-
Second difference:
So, the second difference is constant and equal to 6, which indicates that this sequence is quadratic.
2. Fill in multiples of and subtract:
Here, we compare the original sequence with the sequence of squares of integers:
- values for are: , , , , , .
Now subtract the multiples of from the original sequence:
3. General formula:
Since the second difference is constant, the sequence is quadratic. The general form of a quadratic sequence is: Using the known values from the original sequence and solving for , , and , we can derive the formula.
If we substitute values from the original sequence and solve, we get:
So, the nth term formula is:
Do you need any further clarifications? Here are a few related questions you might consider:
- How are first and second differences used to identify the type of sequence?
- Can you describe how to derive the general form of a quadratic sequence?
- Why is the second difference constant for quadratic sequences?
- What other types of sequences can be identified using difference methods?
- How would the approach differ for a cubic sequence?
Tip: For quadratic sequences, the second difference is always constant, which helps quickly identify the nature of the sequence.
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Math Problem Analysis
Mathematical Concepts
Quadratic Sequences
First Difference
Second Difference
Formulas
a_n = An^2 + Bn + C
nth term formula of a quadratic sequence
Theorems
Quadratic Difference Method
Suitable Grade Level
Grades 9-10
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