Respuesta :
Answer:
[tex]\huge\boxed{a_n=1\cdot(-2)^{n-1}}[/tex]
Step-by-step explanation:
[tex]a_n=a_1r^{n-1}[/tex]
We have the geometric sequence: 1, -2, 4, -8, ...
Find the common ratio:
[tex]r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=...=\dfrac{a_n}{a_{n-1}}[/tex]
[tex]a_1=1,\ a_2=-2[/tex]
substitute:
[tex]r=\dfrac{-2}{1}=-2[/tex]
Substitute
[tex]a_1=1;\ r=-1[/tex]
to
[tex]a_n=a_1r^{n-1}[/tex]
[tex]a_n=1\cdot(-2)^{n-1}[/tex]
Answer:
[tex]a_{n}=1 (-2)^{n-1}[/tex]
Step-by-step explanation:
Hi there!
Let's do it,
1)[tex]1,-2,4,-8[/tex]
First let's find the q, the common ratio by dividing the second term by the anterior one.
-2:1 =-2
4:-2=-2
So the common ratio is -2. Plugging in on the General formula:
[tex]a_{n}=a_{1}q^{n-1}[/tex]
[tex]a_{n}=1 (-2)^{n-1}[/tex]
