How do you find #a_25# given #a_n=(-1)^n(3n-2)#?

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Answer 1 · 2017-08-22T14:51:06

See a solution process below:

Substitute #color(red)(25)# for every occurrence of #color(red)(n)# in #a_n#:

#a_color(red)(n) = (-1)^color(red)(n)(3color(red)(n) - 2)# becomes:

#a_color(red)(25) = (-1)^color(red)(25)((3 * color(red)(25)) - 2)#

#-1^("even")# power = 1 (#-1^2 = 1#; #-1^4 = 1#)

#-1^("odd")# power = -1 (#-1^3 = -1#; #-1^5 = 1#)

Therefore: #-1^25 = -1#

So we can rewrite as:

#a_color(red)(25) = -1((3 * color(red)(25)) - 2)#

#a_color(red)(25) = -1(75 - 2)#

#a_color(red)(25) = -1(73)#

#a_color(red)(25) = -73#