Find the first, fourth, and eighth terms of the sequence.

1 answer:

6 0

$a(n)=-3\cdot2^{n}-1\\\\a_1=-3\cdot2^1-1=-3\cdot2-1=-6-1=-7\\a_4=-3\cdot2^4-1=-3\cdot16-1=-48-1=-49\\a_8=-3\cdot2^8-1=-3\cdot256-1=-768-1=-769\\\\if\ a(n)=-3\cdot2^{n-1}\ then\\\\a_1=-3\cdot2^{1-1}=-3\cdot2^0=-3\cdot1=-3\\a_4=-3\cdot2^{4-1}=-3\cdot2^3=-3\cdot8=-24\\a_8=-3\cdot2^{8-1}=-3\cdot2^7=-3\cdot128=-384$

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