How do you write the first six terms of the sequence #a_n=2^(n+3)#?

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Answer 1 · 2016-11-13T09:26:46

16, 32, 64, 128, 256, 512

You can substitute the natural numbers from 1 to 6 in the expression #2^(n+3)# to get:

#a_color(red)1=2^(color(red)1+3)=2^4=16#

#a_color(red)2=2^(color(red)2+3)=2^5=32#

#a_color(red)3=2^(color(red)3+3)=2^6=64#

#a_color(red)4=2^(color(red)4+3)=2^7=128#

#a_color(red)5=2^(color(red)5+3)=2^8=256#

#a_color(red)6=2^(color(red)6+3)=2^4=512#