Question

Find the first six terms of the sequence.

Answer 1

4, 8, 16, 32, 64, 128

hope this helps :P

Answer 2

Answer: 4,8,16,32,64,128

Step-by-step explanation:-

Given: The first term of sequence=$a_1=4$

The given explicit rule for nth term is

$a_n=2\cdot a_{n-1}$

Now, the second term =$a_2=2\cdot a_{2-1}=2\cdot a_{1}=2(4)=8$

The third term =$a_3=2\cdot a_{3-1}=2\cdot a_{2}=2(8)=16$

The fourth term =$a_4=2\cdot a_{4-1}=2\cdot a_{3}=2(16)=32$

The fifth term =$a_5=2\cdot a_{5-1}=2\cdot a_{4}=2(32)=64$

The sixth term =$a_6=2\cdot a_{6-1}=2\cdot a_{5}=2(64)=128$

Hence, the  first six terms of the sequence are 4,8,16,32,64,128