anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
need you amistre
amistre64
  • amistre64
the top is even sums right?
amistre64
  • amistre64
the bottom is cycling thru odds

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amistre64
  • amistre64
....same problem btm? ;)
anonymous
  • anonymous
yes
amistre64
  • amistre64
k, give me a minute :)
anonymous
  • anonymous
yes top and bottom are AP's
anonymous
  • anonymous
k thanks!
amistre64
  • amistre64
n=1; k=2 n=2; k=4 n=3; k=6 top is 2n
amistre64
  • 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
  • anonymous
yes the general fortmula of the nth term = 2n- top part that is
amistre64
  • amistre64
the bottom is 2n-1 for a
amistre64
  • amistre64
2n --------------- 2n-1.2n+1.2n+3
anonymous
  • anonymous
nth term for btm is(2n-1)(2n+1)(2n+3)
amistre64
  • amistre64
at n=4 8 ------ 7.9.11
amistre64
  • amistre64
times? if thats what the "." means then sure :)
anonymous
  • anonymous
right - but the problem is we need to find the sum of n terms - this is where i got stuck
anonymous
  • anonymous
yes '.' means multiply
amistre64
  • amistre64
well, we have the equation; all we need to do is figure out the summation for it I spose
amistre64
  • amistre64
2n 2n ------------------ = -------------- (2n-1)(2n+1)(2n+3) (4n^2-1)(2n+3)
anonymous
  • anonymous
yeah right - I' must find out what the 'differences ' method is about. I think this stuff is really advanced.
amistre64
  • amistre64
2n ------------------ 8n^3 +12n^2 -2n -6
amistre64
  • amistre64
n ---------------- is what I get for a basis 4n^3 +6n^2 -n -3
anonymous
  • anonymous
yes - thats right
anonymous
  • anonymous
- but where do we go from here?
amistre64
  • amistre64
we could split this by partial fraction decomposition if you think that would help to find seperate summations for each part...
anonymous
  • anonymous
good idea
amistre64
  • amistre64
which the first equation would be more helpful :)
amistre64
  • amistre64
2n A B C ----------------- = ----- + ------ + ------ (2n-1)(2n+1)(2n+3) (2n-1) (2n+1) (2n+3)
anonymous
  • anonymous
yep
amistre64
  • amistre64
2n = A(2n+1)(2n-1) + B(2n-1)(2n+3) + C(2n-1)(2n+1)
amistre64
  • amistre64
rewrite, thats miss typed lol 2n = A(2n+1)(2n+3) + B(2n-1)(2n+3) + C(2n-1)(2n+1)
anonymous
  • anonymous
right 2nd time
amistre64
  • amistre64
when n = 1/2; we get B0 + C0 + A(2)(4) = 1; A = 1/8
amistre64
  • amistre64
when n = -1/2 we get A(0) + C(0) + B(-2)(2) = -1; B=1/4
amistre64
  • amistre64
when n = -3/2; we get B0 + A(0) +C(-4)(-2) = -3; C = -3/8 right?
amistre64
  • amistre64
do you see how I got those numbers? and if so did I do it right :)
anonymous
  • anonymous
yes i understand your working - let me check them
anonymous
  • anonymous
yes -that ok i think
amistre64
  • amistre64
inthe meantime; we will also need to factor out the "2" from the denominators to get the "n" all by itself
amistre64
  • amistre64
1 1 3 ------ - ------- - -------- 8(2n-1) 4(2n+1) 8(2n+3)
amistre64
  • amistre64
1 1 3 --------- - --------- - -------- 16(n-1/2) 8(n+1/2) 16(n+3/2)
amistre64
  • 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
  • amistre64
there is a process to get to the iterations; f(xi) delta(xi)
amistre64
  • 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
  • amistre64
but thats starting to get into the cobwebs in my memory :)
amistre64
  • 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
  • amistre64
any of this sound familiar or reasonable :)
amistre64
  • 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
  • amistre64
9(10)(19) 9(10) -------- - ----- = 1704 -45 = 1659; lol... i guess im rusty :) 6 2
anonymous
  • 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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