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.

- anonymous

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- anonymous

need you amistre

- amistre64

the top is even sums right?

- amistre64

the bottom is cycling thru odds

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## More answers

- amistre64

....same problem btm? ;)

- anonymous

yes

- amistre64

k, give me a minute :)

- anonymous

yes top and bottom are AP's

- anonymous

k thanks!

- amistre64

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

- 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

- anonymous

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

- amistre64

the bottom is 2n-1 for a

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

yes '.' means multiply

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

yes - thats right

- anonymous

- but where do we go from here?

- amistre64

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

- anonymous

good idea

- amistre64

which the first equation would be more helpful :)

- amistre64

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

- anonymous

yep

- amistre64

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

- amistre64

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

- anonymous

right 2nd time

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

yes i understand your working - let me check them

- anonymous

yes -that ok i think

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- 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....

- amistre64

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

- 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

- amistre64

any of this sound familiar or reasonable :)

- 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

- amistre64

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

- 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.

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