Question

MEDALS N BECOMING FAN
1.)Find the first six terms of the sequence.
a1 = -3, an = 2 ● an-1
-6, -12, -24, -48, -96, -192
-3, -6, -12, -24, -48, -96
0, 2, -6, -4, -2, 0
-3, -6, -4, -2, 0, 2

2.)Find an explicit rule for the nth term of the sequence.

2, -8, 32, -128, ...

an = 2 ● 4^(n+1)
an = 2 ● (-4)^n
an = 2 ● 4^(n-1 )
an = 2 ● (-4)^(n-1)

Answer 1

The first sequence, is taking the term before it and multiplying it by 2 every time

Answer 2

starting with -3

Answer 3

\[a _{2}=2 * a_{2-1}\]

a2 would be 2 times a1
a3 would be 2 times a2
a4 would be 2 times a3
.
.
.

Answer 4

-6, -12, -24, -48, -96, -192

Answer 5

but your first term is?

Answer 6

#2 i got an = 2 ● 4^(n-1 )?

Answer 7

-3, -6, -12, -24, -48, -96

Answer 8

sry i meant that 1

Answer 9

yeah that is better

Answer 10

and for 2 u got?

Answer 11

is my answer for 2 correct?

Answer 12

you are multiplying times what every time?

Answer 13

Since they gave you multiple choice, you can try each one and really know nothing..

Answer 14

(the form is a\(_{\rm n}\)=(a\(_{\rm 1}\))•r\(^{\rm n-1}\) )

r - common ratio - (the number by which you multiply each time)

Answer 15

in this case:
2, -8, 32, -128

you are multiplying times 4? That would be close, but not exactly right, because your values are changing signs (they alternate)

Answer 16

2 • what = -8?
-8• what = 32

Answer 17

an = 2 ● (-4)^n @SolomonZelman

Answer 18

ok, i am not going to tell you right or wrong, but this is what I will say

\(a_1=2\cdot (-4)^{1}\)
\(a_1=2\cdot(-4)\)
\(a_1=-8\)

Answer 19

but -8 is the second term, not the first

Answer 20

oh so it has to be this one an = 2 ● (-4)^(n-1)

Answer 21

now,
\(a_n=2\cdot(-4)^{n-1}\)

\(a_2=2\cdot(-4)^{2-1}=2\cdot (-4)^1=-8\)

AND

\(a_1=2\cdot(-4)^{1-1}=2\cdot (-4)^0=2 \cdot 1 = 2\).
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YES IT HAS TO BE \(a_n=2\cdot(-4)^{n-1}\)

Answer 22

u r done with this question:)

Answer 23

thank u!

Answer 24

Anytime!