Find the first four terms of the sequence given by the following. $$ a_{n}=8(2)^{n-1}, n=1,2,3 \ldots $$ $$ \square, \square, \square, \square $$

Question

UpStudy AI · Accepted Answer

The formula for the sequence is given by
$$
a_{n} = 8(2)^{n-1}.
$$

For $$ n = 1 $$:
$$
a_{1} = 8(2)^{1-1} = 8(2)^0 = 8(1) = 8.
$$

For $$ n = 2 $$:
$$
a_{2} = 8(2)^{2-1} = 8(2)^1 = 8(2) = 16.
$$

For $$ n = 3 $$:
$$
a_{3} = 8(2)^{3-1} = 8(2)^2 = 8(4) = 32.
$$

For $$ n = 4 $$:
$$
a_{4} = 8(2)^{4-1} = 8(2)^3 = 8(8) = 64.
$$

Thus, the first four terms of the sequence are $$8, 16, 32, 64$$.
The first four terms of the sequence are 8, 16, 32, and 64.

Find the first four terms of the sequence given by the following. \( a_{n}=8(2)^{n-1}, n=1,2,3 \ldots \) \( \square, \square, \square, \square \)