Question

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.

Answer 1

need you amistre

Answer 2

the top is even sums right?

Answer 3

the bottom is cycling thru odds

Answer 4

....same problem btm? ;)

Answer 5

yes

Answer 6

k, give me a minute :)

Answer 7

yes top and bottom are AP's

Answer 8

k thanks!

Answer 9

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

top is 2n

Answer 10

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

Answer 11

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

Answer 12

the bottom is 2n-1 for a

Answer 13

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

Answer 14

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

Answer 15

at n=4

8
------
7.9.11

Answer 16

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

Answer 17

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

Answer 18

yes '.' means multiply

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

yes - thats right

Answer 25

- but where do we go from here?

Answer 26

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

Answer 27

good idea

Answer 28

which the first equation would be more helpful :)

Answer 29

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

Answer 30

yep

Answer 31

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

Answer 32

rewrite, thats miss typed lol

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

Answer 33

right 2nd time

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

yes i understand your working - let me check them

Answer 39

yes -that ok i think

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

(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....

Answer 46

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

Answer 47

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

Answer 48

any of this sound familiar or reasonable :)

Answer 49

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

Answer 50

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

Answer 51

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.