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

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Answer 1 · 2016-09-17T05:18:40

#3/2#, #1#, #5/6#, #3/4#,#7/10#, #2/3#

#a_n=(n+2)/(2n)#

Plug in n=1, n=2,....n=6

#a_1=(1+2)/(2*1)=3/2#

#a_2=(2+2)/(2*2)=1#

#a_3=(3+2)/(2*3)=5/6#

#a_4=(4+2)/(2*4)=3/4#

#a_5=(5+2)/(2*5)=7/10#

#a_6=(6+2)/(2*6)=2/3#