How do you find the first four terms given #a_1 = 5#; #a_n = 2a_(n-1) + 4#?

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Answer 1 · 2016-08-02T23:01:20

#5,14,32,68#

#{ (a_1 = 5), (a_n = 2a_(n-1)+4) :}#

This formulation describes the sequence #a_1, a_2, a_3,...# recursively.

We find:

#a_1 = 5#

#a_2 = 2a_1+4 = 2(5)+4 = 14#

#a_3 = 2a_2+4 = 2(14)+4 = 32#

#a_4 = 2a_3+4 = 2(32)+4 = 68#

#color(white)()#
Bonus

The general formula for a term of this sequence is described by the formula:

#a_n = 9*2^(n-1)-4#

as we can verify with a proof by induction:

Base case:

#9*2^((1)-1) - 4 = 9*2^0-4 = 9*1-4 = 5 = a_1#

Induction step:

#9*2^((n)-1) - 4 = 2(9*2^((n-1)-1)-4)+4 = 2a_(n-1)+4#