Chapter 12 Sequences and Series - Mixed Review of Problem Solving - Lessons 12.4-12.5 - Page 838: 2c

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From part a, we found that $a_1=1\\a_2=2\\a_3=2^2\\a_4=2^3\\a_5=2^4\\a_6=2^5$ Hence, the explicit rule is $a_n=a^{n-1}$ A recursive rule is $a_1=1\\a_n=2a_{n-1}$ since $a_2=2=2\times1=2a_1\\a_3=2^2=2\times2=2a_2\\a_4=2^3=2\times2^2=2a_3$