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Question: How to find and write the first four terms of the sequences below with the given general term? 1\....

How to find and write the first four terms of the sequences below with the given general term?
1. an=(−1)n+1(n+4){a_n} = {\left( { - 1} \right)^{n + 1}}\left( {n + 4} \right)
2. an=(−1)n−1n2{a_n} = {\left( { - 1} \right)^{n - 1}}{n^2}

Explanation

Solution

Here, we are required to find the first four terms of the given sequences whose general terms are given to us. Thus, we will simply substitute the value of n=1,2,3,4n = 1,2,3,4 respectively to find the first four terms of the given two sequences. Thus, with the help of general terms, we can find any required term of the given sequence.

Complete step by step solution:
In order to find the first four terms when we are given the general term of a sequence, we simply substitute the value of nn by the number or the term which we are required to find.
an=(−1)n+1(n+4){a_n} = {\left( { - 1} \right)^{n + 1}}\left( {n + 4} \right)
In this general term, when we substitute n=1n = 1, we get,
a1=(−1)1+1(1+4)=(−1)2(5)=5{a_1} = {\left( { - 1} \right)^{1 + 1}}\left( {1 + 4} \right) = {\left( { - 1} \right)^2}\left( 5 \right) = 5
Hence, the first term is 5
Similarly, if we substitute n=2,3,4n = 2,3,4 respectively, we get,
Second term, a2=(−1)2+1(2+4)=(−1)3(6)=−6{a_2} = {\left( { - 1} \right)^{2 + 1}}\left( {2 + 4} \right) = {\left( { - 1} \right)^3}\left( 6 \right) = - 6
Third term, a3=(−1)3+1(3+4)=(−1)4(7)=7{a_3} = {\left( { - 1} \right)^{3 + 1}}\left( {3 + 4} \right) = {\left( { - 1} \right)^4}\left( 7 \right) = 7
Fourth term, a4=(−1)4+1(4+4)=(−1)5(8)=−8{a_4} = {\left( { - 1} \right)^{4 + 1}}\left( {4 + 4} \right) = {\left( { - 1} \right)^5}\left( 8 \right) = - 8
Hence, the first four terms of the given sequence an=(−1)n+1(n+4){a_n} = {\left( { - 1} \right)^{n + 1}}\left( {n + 4} \right) are: 5,−6,7,−85, - 6,7, - 8
Where, due to (−1)n+1{\left( { - 1} \right)^{n + 1}}, all the odd terms were positive and the even terms were negative.
an=(−1)n−1n2{a_n} = {\left( { - 1} \right)^{n - 1}}{n^2}
In this general term, when we substitute n=1n = 1, we get,
a1=(−1)1−1(1)2=(−1)0(1)=1{a_1} = {\left( { - 1} \right)^{1 - 1}}{\left( 1 \right)^2} = {\left( { - 1} \right)^0}\left( 1 \right) = 1 (Because (−1)0=1{\left( { - 1} \right)^0} = 1)
Hence, the first term is 1
Similarly, if we substitute n=2,3,4n = 2,3,4 respectively, we get,
Second term, a2=(−1)2−1(2)2=(−1)1(4)=−4{a_2} = {\left( { - 1} \right)^{2 - 1}}{\left( 2 \right)^2} = {\left( { - 1} \right)^1}\left( 4 \right) = - 4
Third term, a3=(−1)3−1(3)2=(−1)2(9)=9{a_3} = {\left( { - 1} \right)^{3 - 1}}{\left( 3 \right)^2} = {\left( { - 1} \right)^2}\left( 9 \right) = 9
Fourth term, a4=(−1)4−1(4)2=(−1)3(16)=−16{a_4} = {\left( { - 1} \right)^{4 - 1}}{\left( 4 \right)^2} = {\left( { - 1} \right)^3}\left( {16} \right) = - 16
Hence, the first four terms of the given sequence an=(−1)n−1n2{a_n} = {\left( { - 1} \right)^{n - 1}}{n^2} are: 1,−4,9,−161, - 4,9, - 16
Where, due to (−1)n+1{\left( { - 1} \right)^{n + 1}}, all the odd terms were positive and the even terms were negative.
Hence, this is the required answer.

Note:
In this question, the given sequences are not considered as an arithmetic progression because their common differences are not the same. An Arithmetic Progression is a sequence of numbers such that the difference between any term and its preceding term is constant. This difference is known as the common difference of an Arithmetic Progression (AP). A real-life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.