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

Nov 13, 2016

16, 32, 64, 128, 256, 512

#### Explanation:

You can substitute the natural numbers from 1 to 6 in the expression ${2}^{n + 3}$ to get:

${a}_{\textcolor{red}{1}} = {2}^{\textcolor{red}{1} + 3} = {2}^{4} = 16$

${a}_{\textcolor{red}{2}} = {2}^{\textcolor{red}{2} + 3} = {2}^{5} = 32$

${a}_{\textcolor{red}{3}} = {2}^{\textcolor{red}{3} + 3} = {2}^{6} = 64$

${a}_{\textcolor{red}{4}} = {2}^{\textcolor{red}{4} + 3} = {2}^{7} = 128$

${a}_{\textcolor{red}{5}} = {2}^{\textcolor{red}{5} + 3} = {2}^{8} = 256$

${a}_{\textcolor{red}{6}} = {2}^{\textcolor{red}{6} + 3} = {2}^{4} = 512$