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In this final part of Exercise 8.1, we push our skills to the limit by combining explicit backwards-solving and multi-layered recursive formulas. [span_2](start_span)[span_2](end_span)
Remember, while explicit formulas use position \(n\), recursive formulas force us to use the value of previous terms like \(T_{n-1}\) or even \(T_{n-2}\). Let's master these complex chains! [span_3](start_span)[span_3](end_span)
Remember, while explicit formulas use position \(n\), recursive formulas force us to use the value of previous terms like \(T_{n-1}\) or even \(T_{n-2}\). Let's master these complex chains! [span_3](start_span)[span_3](end_span)
1 Backwards Position Solving
2 Recursive Generation & Checking
3 Multi-Term Recursive Rule
Select the correct option:
In the recursive formula \(t_{n+1} = t_n + 5\), what exactly does the notation \(t_{n+1}\) represent? [span_14](start_span)[span_14](end_span)
Concept:
1
Subscripts in sequence notation strictly denote the position index. [span_15](start_span)[span_15](end_span)
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Hint:
1
If \(n=4\) (the fourth term), what position is \(n+1\)?
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Mistakes to Avoid:
!
Confusing subscript with regular addition: \(t_{n+1}\) is NOT the same as \(t_n + 1\). The first asks for the next number in the list. The second takes the current number and adds 1 to its value.
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Solution:
In the formula \(t_{n+1} = t_n + 5\), the notation \(t_n\) represents the current term, and \(t_{n+1}\) represents the position immediately following it (the next term). [span_16](start_span)[span_16](end_span)
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Select the correct option:
If \(T_1 = 1\), \(T_2 = 3\), and a recursive sequence follows the rule \(T_n = T_{n-1} \times T_{n-2}\), what is the value of the 4th term (\(T_4\))?
Concept:
1
A recursive rule can combine previous terms using multiplication instead of addition.
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Hint:
1
First find \(T_3\) by multiplying \(T_2\) and \(T_1\).
2
Then find \(T_4\) by multiplying \(T_3\) and \(T_2\).
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Mistakes to Avoid:
!
Skipping terms: You cannot jump straight to \(T_4\). You MUST calculate \(T_3\) first to use it in the formula for \(T_4\).
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Solution:
Given: \(T_1 = 1\), \(T_2 = 3\), and \(T_n = T_{n-1} \times T_{n-2}\)
\(T_3 = T_2 \times T_1 = 3 \times 1 = 3\)
\(T_4 = T_3 \times T_2 = 3 \times 3 = 9\)
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Select the correct option:
If a sequence is defined recursively as \(a_1 = 10\) and \(a_n = a_{n-1} - 2\), what is the value of \(a_3\)?
Concept:
1
A recursive rule can dictate subtraction, meaning the sequence will continually decrease.
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Hint:
1
Calculate \(a_2\) by subtracting 2 from \(a_1\).
2
Then subtract 2 from \(a_2\) to find \(a_3\).
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Mistakes to Avoid:
!
Treating it explicitly: Do not calculate \(10 - 2(3) = 4\). This is a recursive formula, not \(10 - 2n\).
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Solution:
Given: \(a_1 = 10\) and \(a_n = a_{n-1} - 2\)
\(a_2 = a_1 - 2 = 10 - 2 = 8\)
\(a_3 = a_2 - 2 = 8 - 2 = 6\)
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Select the correct option:
If a sequence is built with the rule \(t_n = t_{n-1} + 4\) (starting with \(t_1 = 1\)), how can we easily check if 45 is a term? [span_17](start_span)[span_17](end_span)
Concept:
1
While recursive rules build step-by-step, simple linear ones (like adding a constant 4) can be translated into explicit rules. [span_18](start_span)[span_18](end_span)
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Hint:
1
If a pattern is \(1, 5, 9, 13\dots\) (adding 4), the explicit rule is \(4n - 3\).
2
We can set \(4n - 3 = 45\) and solve for \(n\) much faster than writing out terms! [span_19](start_span)[span_19](end_span)
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Mistakes to Avoid:
!
Assuming manual addition is the only way: You CAN add 4 manually until you hit 45, but it is prone to error and very slow for large numbers. Converting to explicit is superior.
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Solution:
As shown in the textbook exercises, when a recursive sequence involves adding a constant value, it forms an Arithmetic Progression (AP).
We can convert the recursive rule into an explicit formula and solve for the position \(n\) algebraically. [span_20](start_span)[span_20](end_span)
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Select the correct option:
In a "Tribonacci" sequence (where each term is the sum of the three previous terms), if the first three terms are 0, 1, and 1, what is the 5th term? [span_21](start_span)[span_21](end_span)
Concept:
1
A rule that sums the previous three terms is written as \(T_n = T_{n-1} + T_{n-2} + T_{n-3}\). [span_22](start_span)[span_22](end_span)
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Hint:
1
Find the 4th term: \(0 + 1 + 1 = 2\).
2
Find the 5th term by adding the 2nd, 3rd, and 4th terms. [span_23](start_span)[span_23](end_span)
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Mistakes to Avoid:
!
Including the 1st term again: When finding the 5th term, do NOT add the 0 again. You only sum the three most recent terms (the 2nd, 3rd, and 4th).
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Solution:
Given: \(T_1 = 0\), \(T_2 = 1\), \(T_3 = 1\)
\(T_4 = T_3 + T_2 + T_1 = 1 + 1 + 0 = 2\)
\(T_5 = T_4 + T_3 + T_2 = 2 + 1 + 1 = 4\)
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★ Complex Recursive Equations
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