## anonymous one year ago Find the first four terms of a squence using the recursive definition. f(1)=6, f(n)=f(n-1)-5

1. amistre64

what does the definition tell us?

2. anonymous

Using a process that can be repeated. So like the arithmetic sequence. The arithmetic sequence is adding a constant of the previous term.

3. amistre64

ok, in this case, what are we 'adding' to one term to make the next term? any ideas?

4. amistre64

or if we simply let n={1,2,3,4} and the rule defined, we get: f(1) = 6 f(2) = f(1) -5 f(3) = f(2) -5 f(4) = f(3) -5

5. amistre64

so, what would our 4 terms be?

6. anonymous

So since its asking for the first four terms wouldnt it be f(1)=6 f(n)=f(n-1)-5 f(2)=f(2-1)-5=f(1)+1= 6+1=7 so 7 would be one of the four terms but I dont know how to get the rest.

7. amistre64

f(2) = f(2-1) - 5 f(2) = f(1) - 5 f(2) = 6 - 5 = 1, not 7 so, f(1) = 6, f(2) = 1 ---------------------- now repeat with n=3 f(3) = f(3-1) - 5 f(3) = f(2) - 5 f(3) = 1 - 5 = -4 so, f(1) = 6, f(2) = 1, f(3) = -4 what is f(4) ?

8. amistre64

as we find each new term, we can use it to find the next one .... hence the naming it: 'recursive'

9. amistre64

f(4) = f(4-1) - 5 f(4) = f(3) - 5 ^^^ but we know f(3) = -1 f(4) = -1 - 5

10. amistre64

pfft, f(3) = -4 .... these tired old eyes play tricks on me

11. amistre64

can you tell me how we are working this? its basic to me so i cant really see the difficulty. you have to tell me what it is that is confusing you

12. anonymous

Oka, I understand but im confused on f(4)

13. amistre64

ok, tell me how you are looking at f(4) show me your working

14. amistre64

if you know how to get from f(1) to f(2), then the process does not change. f(2) allows us to get f(3), and f(3) allows us to get f(4). typing errors aside, its pretty repetitive.

15. anonymous

So f(4)=-6?

16. amistre64

f(4) = f(3) - 5 f(3) = -4 [i mistyped it before as -1] f(4) = -4 -5

17. amistre64

another way to look at it, just subtract 5 from the setup before ... f(1) = 6 f(2) = 6-5 f(3) = 6-5-5 f(4) = 6-5-5-5

18. anonymous

So f(4)=-9

19. amistre64

yes

20. amistre64

now, what are our 4 terms?

21. anonymous

1,-5,-4,-9

22. amistre64

f(1) = 6 f(2) = 6-5 = 1 f(3) = 6-5-5 = -4 f(4) = 6-5-5-5 = -9

23. anonymous

oh im sorry I didnt mean to write -5 but thank you so much

24. amistre64

$\begin{pmatrix}n\\f(n)\end{pmatrix}=\begin{pmatrix}1&2&3&4\\6&1&-4&-5\end{pmatrix}$

25. amistre64

good luck :)

26. amistre64

yeah, 6, 1 -4, -9 math and typing dont mix that well

27. anonymous

I agree on that ,Do you think you can help me with a couple more?

28. amistre64

maybe one more; its late (1235) and work comes early in the morning

29. anonymous

Okay thank you. f(1)=2,f(n)=-3f(n-1)+[f(n-1)]^2

30. amistre64

well this ones just a exercise in plugging in the value ... show me your work for n=2

31. amistre64

dunno if it helps any to rewrite it by factoring: f(n) = f(n-1) [f(n-1) -3] your call

32. anonymous

Wait okay so do I plug 2 in for n?

33. amistre64

ideally yes, since 2 is the number after 1.

34. anonymous

Okay so I have f(2)=-3f(2-1)+[f(2-1)]^2

35. amistre64

good, and since 2-1 = 1 f(2) = -3 f(1) + [f(1)]^2 and what does f(1) equal?

36. anonymous

Im not sure in what to do after

37. amistre64

you replace f(1) with what its vale is, and work the math

38. amistre64

*value

39. anonymous

So would Imultiply -3 by 1? =-3

40. anonymous

Or where would I get the value from?

41. amistre64

f(1) has been defined for you already in your setup f(1)=2 <------ f(n)=-3f(n-1)+[f(n-1)]^2 ----------------- f(2) = -3 f(1) + [f(1)]^2 f(2) = -3(2) + 2^2 = -6+4 = -2 f(2) = -2 ------------------ we know f(2) now, so let n=3 and work the process again

42. anonymous

Okay thank you so much

43. amistre64

youre welcome .... once we know a new term, we can use it to find the next new term f(3) = -3 f(2) + [f(2)]^2 f(3) = -3(-2) + [-2]^2 = 6+4 f(3) = 10 ---------------- f(4) = -3 f(3) + [f(3)]^2 f(4) = -3(10) + 10^2 = 100-30 etc ....