What is the common ratio of the geometric sequence $ 1,4,16,64,..... $ ?

Hint: From the given series of geometric sequences, we find the general term of the series. We find the formula for $ {{t}_{n}} $ , the $ {{n}^{th}} $ term of the series. From the given sequence we find the common ratio which is the ratio between two consecutive terms. We put the values to get the formula for the general term $ {{t}_{n}} $ . Then we put the value of consecutive natural numbers for $ n $ .

Complete step-by-step answer:
We have been given a series of geometric sequence which is $ 1,4,16,64,..... $
We express the geometric sequence in its general form.
We express the terms as $ {{t}_{n}} $ , the $ {{n}^{th}} $ term of the series.
The first term be $ {{t}_{1}} $ and the common ratio be $ r $ where $ r=\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{{t}_{4}}}{{{t}_{3}}} $ .
We can express the general term $ {{t}_{n}} $ based on the first term and the common ratio.
The formula being $ {{t}_{n}}={{t}_{1}}{{r}^{n-1}} $ .
The first term is 1. So, $ {{t}_{1}}=1 $ . The common ratio is $ r=\dfrac{4}{1}=\dfrac{16}{4}=\dfrac{64}{16}=4 $ .
The common ratio of the geometric sequence $ 1,4,16,64,..... $ is 4.
So, the correct answer is “4”.

Note: We put the values of $ {{t}_{1}} $ and $ r $ to find the general form.
We express general term $ {{t}_{n}} $ as $ {{t}_{n}}={{t}_{1}}{{r}^{n-1}}={{4}^{n-1}} $ .
Now we place consecutive natural numbers for $ n $ as $ 1,2,3,4,... $ to get the sequence as
\[{{4}^{1-1}},{{4}^{2-1}},{{4}^{3-1}},{{4}^{4-1}},......=1,4,16,64,.....\] .
The value of $ \left| r \right|>1 $ for which the sum of the first n terms of the G.P. is $ {{S}_{n}}={{t}_{1}}\dfrac{{{r}^{n}}-1}{r-1}=\dfrac{{{4}^{n}}-1}{3} $ .