How do you find #a_13# given #a_n=(4n^2-n+3)/(n(n-1)(n+2))#?

Question

1403 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-21T19:25:46

Plug in #13# for #n#

#a_13 = 111/286 = 0.3overline(881118)#

#a_n=(4n^2-n+3)/(n(n-1)(n-2))#

#therefore a_13 = (4(13)^2-13+3)/(13(13-1)(13-2))#

#= (4*169-10)/(156(11))#

#= 666/1716#

#= 111/286#

#= 0.3overline(881118)#

Final Answer