Question

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

Answer 1

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

Complete step-by-step answer:
The given ${n^{th}}$ term of an arithmetic sequence is ${a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}$
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 $n = 1,2,3,4$ in the ${n^{th}}$ term of the sequence.
First, we shall apply $n = 1$ in ${a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}$
Thus, we have ${a_1} = {\left( { - 1} \right)^{1 - 1}} \times {2^{1 + 1}}$
$ \Rightarrow {a_1} = {\left( { - 1} \right)^0} \times {2^2}$
$ \Rightarrow {a_1} = 4$
Now, we shall substitute $n = 2$ in ${a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}$
Thus, we have ${a_2} = {\left( { - 1} \right)^{2 - 1}} \times {2^{2 + 1}}$
$ \Rightarrow {a_2} = {\left( { - 1} \right)^1} \times {2^3}$
$ \Rightarrow {a_2} = - 1 \times 8$
$ \Rightarrow {a_2} = - 8$
Now, we shall substitute $n = 3$ in ${a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}$
Thus, we have ${a_3} = {\left( { - 1} \right)^{3 - 1}} \times {2^{3 + 1}}$
$ \Rightarrow {a_3} = {\left( { - 1} \right)^2} \times {2^4}$
$ \Rightarrow {a_3} = 16$
Now, we shall substitute $n = 4$ in ${a_n} = {\left( { - 1} \right)^{n - 1}} \times {2^{n + 1}}$
Thus, we have ${a_4} = {\left( { - 1} \right)^{4 - 1}} \times {2^{4 + 1}}$
$ \Rightarrow {a_4} = {\left( { - 1} \right)^3} \times {2^5}$
\[ \Rightarrow {a_4} = - 1 \times 32\]
$ \Rightarrow {a_4} = - 32$
Hence, we got the required first four terms of the sequence and they are $4, - 8,16, - 32$

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