How do you write the first five terms of the sequence #a_n=(6n)/(3n^2-1)#?

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Answer 1 · 2017-04-24T07:07:26

First five terms of the sequence are #{3,12/11,9/13,24/47,15/37}#

We can find the first five terms by putting value of #n# from #1# to #5#.

As #a_n=(6n)/(3n^2-1)#,

#a_1=(6xx1)/(3xx1^2-1)=6/(3-1)=6/2=3#

#a_2=(6xx2)/(3xx2^2-1)=12/(12-1)=12/11#

#a_3=(6xx3)/(3xx3^2-1)=18/(27-1)=18/26=9/13#

#a_4=(6xx4)/(3xx4^2-1)=24/(48-1)=24/47#

#a_5=(6xx5)/(3xx5^2-1)=30/(75-1)=30/74=15/37#

Hence, first five terms of the sequence are #{3,12/11,9/13,24/47,15/37}#