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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.
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.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the top is even sums right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the bottom is cycling thru odds

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0....same problem btm? ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0k, give me a minute :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes top and bottom are AP's

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0n=1; k=2 n=2; k=4 n=3; k=6 top is 2n

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0n=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
 5 years ago
Best ResponseYou've already chosen the best response.0yes the general fortmula of the nth term = 2n top part that is

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the bottom is 2n1 for a

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02n  2n1.2n+1.2n+3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nth term for btm is(2n1)(2n+1)(2n+3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at n=4 8  7.9.11

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0times? if thats what the "." means then sure :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right  but the problem is we need to find the sum of n terms  this is where i got stuck

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes '.' means multiply

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, we have the equation; all we need to do is figure out the summation for it I spose

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02n 2n  =  (2n1)(2n+1)(2n+3) (4n^21)(2n+3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah right  I' must find out what the 'differences ' method is about. I think this stuff is really advanced.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02n  8n^3 +12n^2 2n 6

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0n  is what I get for a basis 4n^3 +6n^2 n 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0 but where do we go from here?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we could split this by partial fraction decomposition if you think that would help to find seperate summations for each part...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0which the first equation would be more helpful :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02n A B C  =  +  +  (2n1)(2n+1)(2n+3) (2n1) (2n+1) (2n+3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02n = A(2n+1)(2n1) + B(2n1)(2n+3) + C(2n1)(2n+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0rewrite, thats miss typed lol 2n = A(2n+1)(2n+3) + B(2n1)(2n+3) + C(2n1)(2n+1)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when n = 1/2; we get B0 + C0 + A(2)(4) = 1; A = 1/8

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when n = 1/2 we get A(0) + C(0) + B(2)(2) = 1; B=1/4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when n = 3/2; we get B0 + A(0) +C(4)(2) = 3; C = 3/8 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do you see how I got those numbers? and if so did I do it right :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i understand your working  let me check them

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0inthe meantime; we will also need to factor out the "2" from the denominators to get the "n" all by itself

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01 1 3      8(2n1) 4(2n+1) 8(2n+3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01 1 3      16(n1/2) 8(n+1/2) 16(n+3/2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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
 5 years ago
Best ResponseYou've already chosen the best response.0there is a process to get to the iterations; f(xi) delta(xi)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(ba)/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
 5 years ago
Best ResponseYou've already chosen the best response.0but thats starting to get into the cobwebs in my memory :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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) = n1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0any of this sound familiar or reasonable :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets test my memory here: suppose we want to sum the numbers 19; that should equal 45 the equation would simply be "n" n(n1) = n^2 n our summation would amount to: n(n+1)(2n+1) n(n+1)    if my head is right :) 6 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.09(10)(19) 9(10)    = 1704 45 = 1659; lol... i guess im rusty :) 6 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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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