How do you find #a_13# given #a_n=(4n^2-n+3)/(n(n-1)(n+2))#? Precalculus Sequences Infinite Sequences 1 Answer John D. May 21, 2017 Plug in #13# for #n# #a_13 = 111/286 = 0.3overline(881118)# Explanation: #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 Answer link Related questions What is a sequence? How does the Fibonacci sequence relate to Pascal's triangle? What is the Fibonacci sequence? How do I find the #n#th term of the Fibonacci sequence? How do you find the general term for a sequence? How do find the #n#th term in a sequence? What is the golden ratio? How does the golden ratio relate to the Fibonacci sequence? How do you determine if -10,20,-40,80 is an arithmetic or geometric sequence? How do you determine if 15,-5,-25,-45 is an arithmetic or geometric sequence? See all questions in Infinite Sequences Impact of this question 1328 views around the world You can reuse this answer Creative Commons License