Expert answers to "Find a recurrence relation satisfied by this sequence...."

ossidianaZ

Answered question

2021-09-13

Find a recurrence relation satisfied by this sequence.

a_{n} = n + {(- 1)}^{n}

Answer & Explanation

Aubree Mcintyre

Skilled2021-09-14Added 73 answers

Given:
$a_{n} = n + {(- 1)}^{n}$
Let us first determine the first term by replacing n in the given expression for $a_{n}$ by 0:
$a_{0} = 0 + {(- 1)}^{0} = 0 + 1 = 1$
Let us similarly determine the next few terms as well:
$a_{1} = 1 + {(- 1)}^{1} = 1 - 1 = 0 = a_{0} - 1$
$a_{2} = 2 + {(- 1)}^{2} = 2 + 1 = 3 = a_{1} + 3$
$a_{3} = 3 + {(- 1)}^{3} = 3 - 1 = 2 = a_{2} - 1$
$a_{4} = 4 + {(- 1)}^{4} = 4 + 1 = 5 = a_{3} + 3$
$a_{5} = 5 + {(- 1)}^{5} = 5 - 1 = 4 = a_{4} - 1$
$a_{6} = 6 + {(- 1)}^{6} = 6 + 1 = 7 = a_{5} + 3$
We note that each term is the previous term increased by 3 if n is even:
If n even: $a_{n} = a_{n - 1} + 3$
We note that each term is the previous term decreased by 1 if n is odd:
If n odd: $a_{n} = a_{n - 1} - 1$
Thus a recurrence relation for $a_{n}$ is then:
$a_{0} = 1$
$a_{n} = {\begin{cases} a_{n - 1} + 3 & i f n e v e n \\ a_{n - 1} - 1 & i f n o d d \end{cases}$
Note: There are infinitely many different recurence relations that satisfy any sequence.

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