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Chapter 9: Problem 19

Write the first four terms of the sequence \(\left\\{a_{n}\right\\}\) defined by the following recurrence relations. $$a_{n+1}=3 a_{n}-12 ; \quad a_{1}=10$$

Short Answer

Expert verified
Answer: The first four terms of the sequence are $$a_{1}=10$$, $$a_{2}=18$$, $$a_{3}=42$$, and $$a_{4}=114$$.

Step by step solution

01

Understanding the given recurrence relation

The recurrence relation is $$a_{n+1}=3a_{n}-12$$, which means the (n+1)th term of the sequence can be found by multiplying the n-th term by 3 and subtracting 12 from the result. Also, we are given the initial term of the sequence, $$a_{1}=10$$.
02

Finding the second term (a_2)

We find the second term by substituting n=1 into the recurrence relation: $$a_{2}=3a_{1}-12=3(10)-12=30-12=18$$ The second term of the sequence, $$a_{2}$$, is 18.
03

Finding the third term (a_3)

We find the third term by substituting n=2 into the recurrence relation: $$a_{3}=3a_{2}-12=3(18)-12=54-12=42$$ The third term of the sequence, $$a_{3}$$, is 42.
04

Finding the fourth term (a_4)

We find the fourth term by substituting n=3 into the recurrence relation: $$a_{4}=3a_{3}-12=3(42)-12=126-12=114$$ The fourth term of the sequence, $$a_{4}$$, is 114. In conclusion, the first four terms of the sequence are $$a_{1}=10$$, $$a_{2}=18$$, $$a_{3}=42$$, and $$a_{4}=114$$.

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Most popular questions from this chapter

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).

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