How do you find #a_11# given #a_n=(4n)/(2n^2-3)#?
1 Answer
Aug 17, 2017
Explanation:
#"to evaluate substitute n = 11 into "a_n#
#rArra_(color(red)(11))=(4xxcolor(red)(11))/(2xx(color(red)(11))^2-3)#
#color(white)(xxxxx)=44/239#
