Question

How do you find the $ {n^{th}} $ term rule for $ 2,8,32,128, \ldots $ ?

Answer 1

Hint: In order to determine the $ {n^{th}} $ term rule for the sequence $ 2,8,32,128, \ldots $ , we will rewrite the numbers in the sequence in terms of odd powers of $ 2 $ . Then, we will determine the $ {n^{th}} $ term $ S\left( n \right) $ of the given sequence by expressing odd components $ 1,3,5,7, \ldots $ in terms of $ {n^{th}} $ term.

Complete step-by-step answer:
Consider the sequence $ 2,8,32,128, \ldots $ .
The numbers in the sequence are powers of $ 2 $ .
Now, let us write each term in the powers of $ 2 $ , we have,
 $ 2 = {2^1} $
 $ 8 = {2^3} $
 $ 32 = {2^5} $
 $ 128 = {2^7} $
Hence, the sequence can be rewritten in the form, $ {2^1},{2^3},{2^5},{2^7} \ldots $ .
Thus, we can say that the numbers in the sequence are odd powers of $ 2 $ . Since, the powers are odd numbers, i.e., $ 1,3,5,7 $ .
 Then, the next upcoming terms will be $ {2^9} = 512 $ , $ {2^{11}} = 2048 $ , so on.
We need to determine the $ {n^{th}} $ term of the sequence.
Thus, we can express odd components $ 1,3,5,7, \ldots $ in terms of $ {n^{th}} $ term as $ 2n - 1 $ .
Let us start with $ n = 1 $ , in order to verify the $ {n^{th}} $ term.
 $ 2 \times 1 - 1 = 1 $
 $ 2 \times 2 - 1 = 3 $
 $ 2 \times 3 - 1 = 5 $
 $ 2 \times 4 - 1 = 7 $ $ \ldots $
And so on.
Hence, the $ {n^{th}} $ term $ S\left( n \right) $ of the given sequence is,
 $ S\left( n \right) = {2^{\left( {2n - 1} \right)}} $
So, the correct answer is “ $ S\left( n \right) = {2^{\left( {2n - 1} \right)}} $ ”.

Note: A sequence can be defined as a function whose domain is either the set of the natural numbers (for infinite sequence), or a set of the first $ n $ natural numbers. Each term in the sequence is called a term or an element or a member.
The $ {n^{th}} $ term is a formula with $ n $ in it which enables us to find any term of a sequence without having to go up from one term to the next. $ n $ stands for the term number so to find the $ {50^{th}} $ term we will just substitute $ 50 $ in the formula in the place of $ n $ .