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Question: Write first \(4\) terms of the sequence. \({a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + ...

Write first 44 terms of the sequence.
an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}

Explanation

Solution

Here, we are given the nth{n^{th}} term of an arithmetic sequence and we are asked to calculate the first four terms of the given sequence. We can note that an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}} contains the variable nn . So, to obtain the first four terms of the given sequence, we need to substitute some values for the variable nn. Since we need to calculate the first four terms, we shall substitute n=1,2,3,4n = 1,2,3,4 in the nth{n^{th}} term of the sequence.

Complete step-by-step answer:
The given nth{n^{th}} term of an arithmetic sequence is an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}
We are asked to obtain the first four terms of the given sequence.
Since we need to calculate the first four terms, we shall substitute n=1,2,3,4n = 1,2,3,4 in the nth{n^{th}} term of the sequence.
First, we shall apply n=1n = 1 in an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}
Thus, we have a1=(1)11×21+1{a_1} = {\left( { - 1} \right)^{1 - 1}} \times {2^{1 + 1}}
a1=(1)0×22\Rightarrow {a_1} = {\left( { - 1} \right)^0} \times {2^2}
a1=4\Rightarrow {a_1} = 4
Now, we shall substitute n=2n = 2 in an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}
Thus, we have a2=(1)21×22+1{a_2} = {\left( { - 1} \right)^{2 - 1}} \times {2^{2 + 1}}
a2=(1)1×23\Rightarrow {a_2} = {\left( { - 1} \right)^1} \times {2^3}
a2=1×8\Rightarrow {a_2} = - 1 \times 8
a2=8\Rightarrow {a_2} = - 8
Now, we shall substitute n=3n = 3 in an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}
Thus, we have a3=(1)31×23+1{a_3} = {\left( { - 1} \right)^{3 - 1}} \times {2^{3 + 1}}
a3=(1)2×24\Rightarrow {a_3} = {\left( { - 1} \right)^2} \times {2^4}
a3=16\Rightarrow {a_3} = 16
Now, we shall substitute n=4n = 4 in an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}
Thus, we have a4=(1)41×24+1{a_4} = {\left( { - 1} \right)^{4 - 1}} \times {2^{4 + 1}}
a4=(1)3×25\Rightarrow {a_4} = {\left( { - 1} \right)^3} \times {2^5}
a4=1×32\Rightarrow {a_4} = - 1 \times 32
a4=32\Rightarrow {a_4} = - 32
Hence, we got the required first four terms of the sequence and they are 4,8,16,324, - 8,16, - 32

Note: We can note that we are given the nth{n^{th}} term of an arithmetic sequence and it is an=(1)n1×2n+1{a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}. To find the first four terms of the sequence, we need to put n=1,2,3,4n = 1,2,3,4 in the nth{n^{th}} term of the sequence. If we are given a sequence, we need to first find the nth{n^{th}} term of the sequence. Then, we need to do the steps as we did earlier.