## anonymous one year ago 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)

1. DanJS

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

2. DanJS

starting with -3

3. DanJS

$a _{2}=2 * a_{2-1}$ a2 would be 2 times a1 a3 would be 2 times a2 a4 would be 2 times a3 . . .

4. anonymous

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

5. SolomonZelman

6. anonymous

#2 i got an = 2 ● 4^(n-1 )?

7. anonymous

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

8. anonymous

sry i meant that 1

9. SolomonZelman

yeah that is better

10. SolomonZelman

and for 2 u got?

11. anonymous

is my answer for 2 correct?

12. SolomonZelman

you are multiplying times what every time?

13. DanJS

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

14. SolomonZelman

(the form is a$$_{\rm n}$$=(a$$_{\rm 1}$$)•r$$^{\rm n-1}$$ ) r - common ratio - (the number by which you multiply each time)

15. SolomonZelman

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)

16. SolomonZelman

2 • what = -8? -8• what = 32

17. anonymous

an = 2 ● (-4)^n @SolomonZelman

18. SolomonZelman

ok, i am not going to tell you right or wrong, but this is what I will say $$a_1=2\cdot (-4)^{1}$$ $$a_1=2\cdot(-4)$$ $$a_1=-8$$

19. SolomonZelman

but -8 is the second term, not the first

20. anonymous

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

21. SolomonZelman

now, $$a_n=2\cdot(-4)^{n-1}$$ $$a_2=2\cdot(-4)^{2-1}=2\cdot (-4)^1=-8$$ AND $$a_1=2\cdot(-4)^{1-1}=2\cdot (-4)^0=2 \cdot 1 = 2$$. ---------------------------------- YES IT HAS TO BE $$a_n=2\cdot(-4)^{n-1}$$

22. SolomonZelman

u r done with this question:)

23. anonymous

thank u!

24. SolomonZelman

Anytime!