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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

need you amistre

- amistre64

the top is even sums right?

- amistre64

the bottom is cycling thru odds

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.