How do you write the nth term rule for the sequence #6,14,22,30,38,...#?

1 Answer
Nov 5, 2016

Answer:

#a_n=8n-2#

Explanation:

There are 2 approaches. The second one is more accepted approach

#color(blue)("First approach - should not be presented as a solution")#

Observe that we have:

#n" sequence build"#

#1" "->6color(white)(1) = 6+0#
#2" "->14 = 6+8#
#3" "->22=6+8+8#
#4" "->30=6+8+8+8#
#5" "->38=6+8+8+8+8#

Now compare the count of 8's to the values of #n#
Notice that we have #n-1# of them

So for any #i^("th")# term we have the term #a_i# where

#a_i=6+8(i-1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Second approach - this is the one to present as a solution")#

Note that #6# is #8-2# so substitute for all the sixes giving:

#n" sequence build"#

#1" "->6color(white)(1) = 8-2#
#2" "->14 = 8+8-2#
#3" "->22=8+8+8-2#
#4" "->30=8+8+8+8-2#
#5" "->38=8+8+8+8+8-2#

The question asks for the #n^("th")# term

So for any #n^("th")# term we have the term #a_n# where

#" "color(blue)(ul(bar( |color(white)(2/2)a_n=(ixx8)-2" "->" "a_n=8n-2color(white)(2/2)|))#