Alvaro Wilkinson

Answered

2022-12-31

How to find and write the first four terms of the sequences below with the given general term?
(1) ${a}_{n}={\left(-1\right)}^{n+1}\left(n+4\right)$
(2) ${a}_{n}={\left(-1\right)}^{n-1}{n}^{2}$

Answer & Explanation

Yosef Santana

Expert

2023-01-01Added 8 answers

${a}_{n}={\left(-1\right)}^{n+1}\left(n+4\right)$
5,6,7, and 8 are the first four terms in the sequence with the general term (n+4).
${\left(-1\right)}^{n+1}$ simply means that terms will be positive when they are odd and negative when they are even.
Therefore, ${a}_{n=1\to 4}={\left(-1\right)}^{n+1}\left(n+4\right)=+5,-6,+7,-8$
${a}_{n}={\left(-1\right)}^{n-1}{n}^{2}$
The first 4 terms of the sequence with general term ${n}^{2}$ are: 1,4,9,16
${\left(-1\right)}^{n-1}$ also means that odd terms will be positive and even terms will be negative. [Since, ${\left(-1\right)}^{0}=1$]
Thus, ${a}_{n=1\to 4}={\left(-1\right)}^{n-1}{n}^{2}=+1,-4,+9,-16$

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