More Worked Examples — Quadratic Sequences
Here are additional examples to deepen your understanding of quadratic sequences and series. These examples include different patterns and contexts, ranging from simple numerical patterns to applied problems.
Example 1: Find the nth term of 4, 9, 16, 25, 36, ...
First differences: 5, 7, 9, 11 → Second differences: 2, 2, 2 (constant). Hence, it’s quadratic.
2a = 2 → a = 1. Assume Tn = n² + b n + c.
For n=1: 1 + b + c = 4 → b + c = 3
For n=2: 4 + 2b + c = 9 → 2b + c = 5
Subtract equations: b = 2, c = 1
Therefore, Tn = n² + 2n + 1 (a perfect square sequence).
Example 2: Find the nth term of 7, 15, 27, 43, 63, ...
First differences: 8, 12, 16, 20 → Second differences: 4, 4, 4 (constant)
2a = 4 → a = 2
Assume Tn = 2n² + b n + c
For n=1: 2 + b + c = 7 → b + c = 5
For n=2: 8 + 2b + c = 15 → 2b + c = 7
Subtract equations: b = 2, c = 3
Therefore, Tn = 2n² + 2n + 3
Example 3: Real-world application — Height of a thrown ball
A ball’s height at 1-second intervals is measured as follows (in meters): 0, 14, 24, 30, 32, 30, 24, 14, 0.
The pattern of heights forms a quadratic sequence because the second differences are constant.
By calculating differences, the formula can be modeled as: h(t) = -2t² + 16t.
Interpretation: The ball reaches its maximum height when t = 4 seconds, and h(4) = 32 m.
Example 4: Find the sum of the first 5 terms of Tn = 3n² - n + 2
a = 3, b = -1, c = 2.
Use formula Sn = a·n(n+1)(2n+1)/6 + b·n(n+1)/2 + c·n
S₅ = 3·(5·6·11)/6 + (-1)·(5·6)/2 + 2·5
S₅ = 165 - 15 + 10 = 160
Example 5: Find the nth term of 10, 17, 28, 43, 62, ...
First differences: 7, 11, 15, 19 → Second differences: 4, 4, 4 (constant)
2a = 4 → a = 2. Assume Tn = 2n² + b n + c
For n=1: 2 + b + c = 10 → b + c = 8
For n=2: 8 + 2b + c = 17 → 2b + c = 9
Subtract equations: b = 1, c = 7
Therefore, Tn = 2n² + n + 7
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