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I'm trying to find a pattern
\[a_{n+2}=\frac{a_n}{(n+2)(n+1)}\]
\[n=0:a_2=\frac{a_0}{2!}\]
\[n=2:a_4=\frac{a_0}{4!}\]
but that pattern changes because the next term is not \[\frac{a_0}{6!}\]
\[n=4:a_6=\frac{a_0}{2\cdot 30}=\]
 one year ago
 one year ago
I'm trying to find a pattern \[a_{n+2}=\frac{a_n}{(n+2)(n+1)}\] \[n=0:a_2=\frac{a_0}{2!}\] \[n=2:a_4=\frac{a_0}{4!}\] but that pattern changes because the next term is not \[\frac{a_0}{6!}\] \[n=4:a_6=\frac{a_0}{2\cdot 30}=\]
 one year ago
 one year ago

This Question is Closed

Algebraic!Best ResponseYou've already chosen the best response.0
?? \[a _{4} = \frac{a _{0} }{ 4! } \]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
this is the one we did yesterday? wasn't it\[a_{n+2}=a_n\frac{1n}{(n+1)(n+2)}\]?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
no this is new problem \[y''=y\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I'll write out what I have so far
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I'll be able to able to help in about 1520 min
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=2}^\infty n(n1)a_nx^{n2}\sum_{n=0}^\infty a_nx^n=0\]
 one year ago

nipunmalhotra93Best ResponseYou've already chosen the best response.0
@MathSofiya how exactly is the pattern changing?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
let's establish that we have the right pattern first... I feel like you may have missed something continue plz @MathSofiya so far so good
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n\sum_{n=0}^\infty a_nx^n=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so yeah, right pattern
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)a_n\right]=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\sum_{n=0}^\infty x^n\left[(n+2)(n+1)a_{n+2}a_n\right]=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[a_{n+2}=\frac{a_n}{(n+1)(n+2)}\]so you were right, just plug in each n and write out \(everything\) explicitly, even multiplication by 1
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
so we established the relationship between \[a_2=\frac{a_0}{2}\] but how do I continue this to the next line? how do I change a_2 to a_0?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ooohh no need to because the next line is going to be zero
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you are not writing it out explicitly enough to see the pattern reduce everything to terms of either a_0 or a_1 the two constants we can represent no more simply
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
now is the next line zero?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
well it was yesterday LOL ....Ok I'll write everything out
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]write everything this way, with each number on the bottom specified
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yesterday we had n1 in the top, that's why
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Haha I'm exercising while you crunch the numbers :P
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{2\cdot 3}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{2\cdot2\cdot3}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=4:a_{4+2}=a_6=\frac {a_4}{30}=\frac{a_0}{2\cdot2\cdot2\cdot3\cdot5}\] \[n=5:a_{5+2}=a_7=\frac {a_5}{42}=\frac{a_1}{2\cdot2\cdot2\cdot3\cdot5\cdot42}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
n=4 is wrong oh well...but I see a pattern now. How do you workout and do math?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
push ups, then run over and check your work real quick :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
haha! That's what I call commitment!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
give me a second to do the problem on paper. It's kinda tedious to type it...one second
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_{0+2}=a_2=\frac {a_0}{2\cdot1}\]\[n=1:a_3=\frac{a_1}{3\cdot2}\]\[n=2:a_4=\frac {a_2}{4\cdot3}=\frac{a_0}{4\cdot3\cdot2\cdot1}\]\[n=3:a_5=\frac {a_3}{5\cdot4}=\frac{a_1}{5\cdot4\cdot3\cdot2}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
write out each number in the denom without simplifying every time
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I cannot stress that enough to find patterns
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[n=0:a_{0+2}=a_2=\frac {a_0}{2}\] \[n=1:a_{1+2}=a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}\] \[n=2:a_{2+2}=a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}\] \[n=3:a_{3+2}=a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}\] Ok I'm finally convinced...had to calculate like 5 times that the denominator is actually 120
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
good, but more importantly what is/are the general pattern(s)?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
again, there are two; one for even n and one for odd n
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
sorry internet is super slow. I'm gonna restart my computer brb
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I'm getting in the shower soon... so what are the two constants that every other constant \(a_n\) can be written in terms of ?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I wanna say \[(n+2)!\] and then sub in n=2k
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
wait...you go shower, and let me work on this
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no, and that would only be evens, and needs a base constant like \(a_0\) write it as\[a_n=?\]and remember this time we do have odd terms
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
allow me to decipher it...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
The denominator for both even and odds is the following...still working on the numerator \[\frac{?}{(n+2)!}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
look at the difference between constants with an even and odd subscript what is the main difference you see>?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
one will always be zero and the other will always be 1. I have come to that conclusion, but now I have to find a way to write it mathematically...I'm gonna cheat and look at yesterdays work
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no wait, you can do it... let me give another hint...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
odd are reducible to being in terms of the constant a_1 evens are reducible to being in terms of a_0 and for the denominator for any constant \(a_n\) what is it? look at the denom for a_2, a_3, a_4, what are they in terms of n ?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
don't get confused with the n=(whatever) thing I mean what is the relation between the n in the subscript \(a_n\) and the number in the denominator?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
hold on I didn't write anything yet
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
sorry It just took me 3 mins to load this page
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no rush, still haven't showered yet
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
When the numerator is \[a_0\] the denominator is even When the numerator is \[a_1\] the denominator is odd
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
by odd I mean 5! 3! and so on
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
exactly, and all even numbers can be written how?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{2k}\] and \[a_{2k+1}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so then what is our formulas? we need one for all \(a_{2k}\) and another for all \(a_2k+1}\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, we need a formula for each one.. what are they ? that's probably the hardest part of the problem in my opinion is finding the pattern
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I would like to put a_{2k} in the numerator and (n+2) in the numerator, but that wouldn't be right. So I have write it in one formula to account for both the even and odd?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I can't write two separate formulas correct? I've gotta write one?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
everything should be written in terms of either a_0 and k, or a_1 and k n is now out of the poicture completely
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
gtg...I know this is the last part, and shouldn't take too long, but I gtg to the gym. i'll be back at 7:30 8pm
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
cool, health makes one think more clearly I believe see ya
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
LOL I hope soo! see ya
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
where is Alan Turing?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I'm reviewing the previous problems we did...and there might be some errors. I not convinced with the way that the final answer was written in k's
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Allow me to restore your faith, though there are other manners of making the notation I'm sure... Paul to the rescue again: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx Notice that since the constants become different depending on evenodd subscripts for many answers we divide the subscripts in the recursion formula for them into terms of 2k and 2k+1. This particular problem you are doing is almost identical to example 1 one this page, with the exception of the minus between y'' and y exchanged for a plus (reading this example you will probably discover the pattern I was asking you to look for). Notice that it is in fact essential that we change the terms from n to k, since we can get no single summation formula in terms of n since the coefficients depend on whether n is odd n. This is not true for all problems with series solutions, but when we need to distinguish between even and odd n the only way to do this is to put them as two separate series in terms of k. Read example 1 carefully to understand the intricacies of the process, and the why we put n in terms of k. Sorry I never returned last night, I was teaching political philosophy. Look forward to finishing the problem with you today if possible. Tag me when you are ready to continue :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I'll be back in like 10 mins, gotta start laundry
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I can't find my laundry card! Ok so I was trying to rewrite the first problem, and got to the part where 1. all even n#'s =0, or all odd "a" subscripts=0 2. We've gotten a pattern that we called \[a_n=\frac{a_0}{2^n(n!)}\] but if we call them "n" for now wouldn't that conflict because Let's say for n=5 \[\frac{a_0}{2^3(3!)}\] which is not true (I'm referring to y'=xy, the first problem)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Well it's what we called the pattern initially before writing it in terms of k's
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ok found the laundry card brb
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
The pattern is not in terms of n like that, if it were it would be\[a_n=\frac{a_0}{2^{n/2}(\frac n2!)}\]for even n subscripts of a, and\[a_n=0\] for all odd n subscripts of a. I think a major problem here is that we are getting confused with the n in the \(n=...\) that we are using to plug in to find the particular constant \(a_n\), and the n in the \(a_n\) in the constant. Those two n's are not equal! What exactly is your argument that in the last problem you are referring too we do no not have that\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Ok that makes sense. I thought the n's were the same. So the n's are just a variable to hold place until we change the argument to k's I guess. wait, what do you mean by the last sentence? so the n=5 statement above is not accurate?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes it is accurate, plugging in n=5 into the recursion formula\[a_{n+1}=\frac{a_{n1}}{n+1}\]gives\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]the \(n=...\) is just what we use to get an expression relating two other constants \(a_{n+ p}\)and \(a_{n+ q}\) where p and q are some number (in this case above p=1 and q=1) so the n we plug in to test is \(not\) the same as the n in the subscripts, because the subscripts in our recursion relation are usually not just \(a_n\) (the same n as we are plugging in for the test) but really \(a_{n+p}\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
We want our expression for each constant in terms of its \(subscript\), not the \(n=...\) we use to test the relation, hence we divide the subscripts into even and odd to make the pattern more clear. Once we notice that by doing that we can get meaningful expressions for each constant a in terms of the subscript (however we write) that is what we use...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
in this case by dividing the subscripts into those of the form 2k and those of the form 2k+1 we get a new way to write each constant in terms of k
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
...and whatever base constant we can't determine, like \(a_0\) which we do not know since we are not given the initial conditions
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
ok that makes more sense now. Let's see if I can apply it to this problem now.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
go for it, I'm gonna get a few empanadas for breakfast :) (sortof like Mexican hotpockets :) brb
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
Let's see here.... \[a_2=\frac{a_0}{2!}\] \[a_3=\frac{a_1}{3!}\] \[a_4=\frac{a_0}{4!}\] \[a_5=\frac{a_1}{5!}\] \[a_2k=\frac{a_0}{k!}\] my way of saying if the subscript is even...
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{2k+1}=\frac{a_1}{k!}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
is it? what is\(a_3\) ?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh my 2k+1=3 2k=2 k=1
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
your patients with me is truly astonishing. I would have strangled the student.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
haha, but I haven't done this in a long time either, so it's a good review for me too.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[a_n=\frac{a_n}{(n+1)(n+2)}\]\[n=0:a_2=\frac {a_0}{2}=\frac{a_0}{2!}\]\[n=1:a_3=\frac {a_1}{6}=\frac{a_1}{3\cdot 2}=\frac{a_1}{3!}\]\[n=2:a_4=\frac {a_2}{12}=\frac{a_0}{4\cdot3\cdot2\cdot1}=\frac{a_0}{4!}\]\[n=3:a_5=\frac {a_3}{20}=\frac{a_3}{10\cdot2}=\frac{a_1}{5\cdot4\cdot3\cdot2\cdot1}=\frac{a_1}{5!}\]now hopefully you have already noticed that the denominator on each \(a_n\) is \(n!\)\[a_n=\frac{a_0}{n!}\]if n is even, and\[a_n=\frac{a_1}{n!}\]if n is odd given that, simply rewrite the subscripts as 2k or 2k+1 depending on whether they are even or odd, and then sub in for n in the general expressions
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yep, I honestly saw that relationship all along, I just didn't know how to express that in k's \[a_{2k}=\frac{a_0}{2k}\] \[a_{2k+1}=\frac{a_0}{2k+1}\] sorry I kinda feel silly now
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you forgot the factorials ;)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
LOL ok let me rewrite them
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
and you forgot that for odd subscripts it is a_1 in the numerator
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{2k}=\frac{a_0}{(2k)!}\] \a_{2k+1}=\frac{a_1}{2k+1}[\] I didn't forgot...I just checking if you were paying attention :P
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{2k+1}=\frac{a_1}{2k+1}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[a_{2k+1}=\frac{a_1}{(2k+1)!}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh wait there is more...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
Much better :) I wanna make a subtle point about the possible values of k we are talking about, but I will save it till the end so as not to confuse you.
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[y=\sum_{n=0}^\infty a_nx^n=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!} \] \[y=\sum_{n=0}^\infty a_nx^n=a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!} \]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you've got the right idea, but we need to use superposition to get the full answer
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
if we accept only one answer or the other we are excluding either all the even or odd exponents, so how can we fix that?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
put them into one formula somehow
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
oh y and c_1 and so on and so forth...hold on
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
\[c_1\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+c_2\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\] huh?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
gosh the sums probably shouldn't be there...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no I was just saying that the c1 you are used to is called a0 in this case, so...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}=0\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[y=a_0\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}+a_1\sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)!}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I just looked at the example in the book Uhm Ohhh,,, but they don't have to be written in c_1 and c_2 right?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
can I put my laundry in the dryer?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
nah, constants are constants, you can call them whatever you want but I just wanted to keep our notation consistent, and since they are equivalent to c1 and c2 (constants that depend on the initial conditions) and yes, you have my permission to do your laundry :P
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so review what we did, look again at example one one here: http://tutorial.math.lamar.edu/Classes/DE/SeriesSolutions.aspx and tell me if you have any questions
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I read it, I don't think I have any questions. 22 hrs later I finally understand it (hopefully I'll be able to recreate it)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
awesome! there is, as I said, a subtle point to be made about the possible values of k (as paul mentions) but since it really doesn't affect this problem I'll leave it for you to read, or we'll cross that bridge when we come to it :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
sounds good...I'll do 2 problems on my own from beginning to end and we'll see if I make error or still struggle on some points here and there...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I do encourage you to read the whole page I linked you to though, to see some of the differences encountered in other problems. Good luck!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
I will. Thanks! quick question, did you make your own empanadas?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
nope, bought 'em at the corner I only cook hamburgers and pasta
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
what? you've been in Mexica for a little while now! Gotta learn to cook like the locals!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
well I make quesadillas, I guess that's a start lol
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.1
yep, work your way up from here :P
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348252365896:dw
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mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348252603116:dw
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mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348252694415:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.0
dw:1348252778070:dw
 one year ago
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