## anonymous 5 years ago i am baffled about how to do the following problem Sum to n terms the following series 2/(1.3.5) + 4/(3.5.7) + 6/(5.7.9) ...... I FOund this in a textbook in a paragraph about summing series by the method of differences.

1. anonymous

need you amistre

2. amistre64

the top is even sums right?

3. amistre64

the bottom is cycling thru odds

4. amistre64

....same problem btm? ;)

5. anonymous

yes

6. amistre64

k, give me a minute :)

7. anonymous

yes top and bottom are AP's

8. anonymous

k thanks!

9. amistre64

n=1; k=2 n=2; k=4 n=3; k=6 top is 2n

10. amistre64

n=1; a=1.b=3.c=5 n=2; a=3.b=5.c=7 n=3; a=5.b=7.c=9 if we solve fo "a" then ther rest are just add ons

11. anonymous

yes the general fortmula of the nth term = 2n- top part that is

12. amistre64

the bottom is 2n-1 for a

13. amistre64

2n --------------- 2n-1.2n+1.2n+3

14. anonymous

nth term for btm is(2n-1)(2n+1)(2n+3)

15. amistre64

at n=4 8 ------ 7.9.11

16. amistre64

times? if thats what the "." means then sure :)

17. anonymous

right - but the problem is we need to find the sum of n terms - this is where i got stuck

18. anonymous

yes '.' means multiply

19. amistre64

well, we have the equation; all we need to do is figure out the summation for it I spose

20. amistre64

2n 2n ------------------ = -------------- (2n-1)(2n+1)(2n+3) (4n^2-1)(2n+3)

21. anonymous

yeah right - I' must find out what the 'differences ' method is about. I think this stuff is really advanced.

22. amistre64

2n ------------------ 8n^3 +12n^2 -2n -6

23. amistre64

n ---------------- is what I get for a basis 4n^3 +6n^2 -n -3

24. anonymous

yes - thats right

25. anonymous

- but where do we go from here?

26. amistre64

we could split this by partial fraction decomposition if you think that would help to find seperate summations for each part...

27. anonymous

good idea

28. amistre64

which the first equation would be more helpful :)

29. amistre64

2n A B C ----------------- = ----- + ------ + ------ (2n-1)(2n+1)(2n+3) (2n-1) (2n+1) (2n+3)

30. anonymous

yep

31. amistre64

2n = A(2n+1)(2n-1) + B(2n-1)(2n+3) + C(2n-1)(2n+1)

32. amistre64

rewrite, thats miss typed lol 2n = A(2n+1)(2n+3) + B(2n-1)(2n+3) + C(2n-1)(2n+1)

33. anonymous

right 2nd time

34. amistre64

when n = 1/2; we get B0 + C0 + A(2)(4) = 1; A = 1/8

35. amistre64

when n = -1/2 we get A(0) + C(0) + B(-2)(2) = -1; B=1/4

36. amistre64

when n = -3/2; we get B0 + A(0) +C(-4)(-2) = -3; C = -3/8 right?

37. amistre64

do you see how I got those numbers? and if so did I do it right :)

38. anonymous

yes i understand your working - let me check them

39. anonymous

yes -that ok i think

40. amistre64

inthe meantime; we will also need to factor out the "2" from the denominators to get the "n" all by itself

41. amistre64

1 1 3 ------ - ------- - -------- 8(2n-1) 4(2n+1) 8(2n+3)

42. amistre64

1 1 3 --------- - --------- - -------- 16(n-1/2) 8(n+1/2) 16(n+3/2)

43. amistre64

Now to sum to the "nth" we can sum each seperately...$[\frac{1}{16}\sum_{}\frac{1}{n-\frac{1}2}{}] -[\frac{1}{8}\sum_{}\frac{1}{(n+\frac{1}{2})}] -[\frac{1}{16}\sum_{}\frac{1}{n+\frac{3}{2}}]$

44. amistre64

there is a process to get to the iterations; f(xi) delta(xi)

45. amistre64

(b-a)/n is the form im familiar with; where a is the leftmost in the interval and b is the rightmost. and n - the number of partitions....

46. amistre64

but thats starting to get into the cobwebs in my memory :)

47. amistre64

if n = the interval 1 to 10 and we partition it out every 1 unitl that would be 9; so I am thinking our delta(xi) = n-1

48. amistre64

any of this sound familiar or reasonable :)

49. amistre64

lets test my memory here: suppose we want to sum the numbers 1-9; that should equal 45 the equation would simply be "n" n(n-1) = n^2 -n our summation would amount to: n(n+1)(2n+1) n(n+1) ------------ - ------- if my head is right :) 6 2

50. amistre64

9(10)(19) 9(10) -------- - ----- = 1704 -45 = 1659; lol... i guess im rusty :) 6 2

51. anonymous

I got the same 3 sum expressions as you did amis but came to a dead end. i believe there must be a different approach to solve this one.