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Find the first five terms of the recursive sequence. Show all your work. an = 3a n-1 - 6 where a1 = 7

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Answer: 7, 15, 39, 111, 327

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Work Shown:

First replace every copy of n with 2


a_n = 3*(a_(n-1)) - 6\\\\a_2 = 3*(a_(2-1)) - 6\\\\a_2 = 3*(a_(1)) - 6\\\\a_2 = 3*(7) - 6\\\\a_2 = 21 - 6\\\\a_2 = 15\\\\

Notice how the second term
a_2 relies on the first term
a_1

Then repeat for n = 3


a_n = 3*(a_(n-1)) - 6\\\\a_3 = 3*(a_(3-1)) - 6\\\\a_3 = 3*(a_(2)) - 6\\\\a_3 = 3*(15) - 6\\\\a_3 = 45 - 6\\\\a_3 = 39\\\\

Same goes with n = 4


a_n = 3*(a_(n-1)) - 6\\\\a_4 = 3*(a_(4-1)) - 6\\\\a_4 = 3*(a_(3)) - 6\\\\a_4 = 3*(39) - 6\\\\a_4 = 117 - 6\\\\a_4 = 111\\\\

Finally plug in n = 5


a_n = 3*(a_(n-1)) - 6\\\\a_5 = 3*(a_(5-1)) - 6\\\\a_5 = 3*(a_(4)) - 6\\\\a_5 = 3*(111) - 6\\\\a_5 = 333 - 6\\\\a_5 = 327\\\\

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We have this summary:


a_1 = 7\\\\a_2 = 15\\\\a_3 = 39\\\\a_4 = 111\\\\a_5 = 327\\\\

The first five terms are: 7, 15, 39, 111, 327

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