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anonymous
 4 years 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 \[\frac{a_0}{6!}\]
\[n=4:a_6=\frac{a_0}{2\cdot 30}=\]
anonymous
 4 years 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 \[\frac{a_0}{6!}\] \[n=4:a_6=\frac{a_0}{2\cdot 30}=\]

This Question is Closed

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no this is new problem \[y''=y\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll write out what I have so far

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I'll be able to able to help in about 1520 min

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=2}^\infty n(n1)a_nx^{n2}\sum_{n=0}^\infty a_nx^n=0\]

nipunmalhotra93
 4 years ago
Best ResponseYou've already chosen the best response.0@MathSofiya how exactly is the pattern changing?

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so yeah, right pattern

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

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

TuringTest
 4 years ago
Best 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ooohh no need to because the next line is going to be zero

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1now is the next line zero?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well it was yesterday LOL ....Ok I'll write everything out

TuringTest
 4 years ago
Best 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yesterday we had n1 in the top, that's why

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Haha I'm exercising while you crunch the numbers :P

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0n=4 is wrong oh well...but I see a pattern now. How do you workout and do math?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1push ups, then run over and check your work real quick :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha! That's what I call commitment!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0give me a second to do the problem on paper. It's kinda tedious to type it...one second

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1write out each number in the denom without simplifying every time

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I cannot stress that enough to find patterns

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1good, but more importantly what is/are the general pattern(s)?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1again, there are two; one for even n and one for odd n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry internet is super slow. I'm gonna restart my computer brb

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I wanna say \[(n+2)!\] and then sub in n=2k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait...you go shower, and let me work on this

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0allow me to decipher it...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The denominator for both even and odds is the following...still working on the numerator \[\frac{?}{(n+2)!}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0one 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no wait, you can do it... let me give another hint...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1odd 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 ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1don'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hold on I didn't write anything yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry It just took me 3 mins to load this page

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no rush, still haven't showered yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When the numerator is \[a_0\] the denominator is even When the numerator is \[a_1\] the denominator is odd

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0by odd I mean 5! 3! and so on

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1exactly, and all even numbers can be written how?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{2k}\] and \[a_{2k+1}\]

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1right, 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I can't write two separate formulas correct? I've gotta write one?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gtg...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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1cool, health makes one think more clearly I believe see ya

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL I hope soo! see ya

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where is Alan Turing?

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Allow 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll be back in like 10 mins, gotta start laundry

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well it's what we called the pattern initially before writing it in terms of k's

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok found the laundry card brb

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1The 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\[n=5:a_6=\frac{a_4}6=\frac{a_0}{2\cdot4\cdot6}=\frac{a_0}{2^3(3!)}\]??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok 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?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yes 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 \(a_{n+p}\)

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1We 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...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1in 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

TuringTest
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok that makes more sense now. Let's see if I can apply it to this problem now.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let'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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{2k+1}=\frac{a_1}{k!}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1is it? what is\(a_3\) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh my 2k+1=3 2k=2 k=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0your patients with me is truly astonishing. I would have strangled the student.

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

TuringTest
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep, 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you forgot the factorials ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0LOL ok let me rewrite them

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1and you forgot that for odd subscripts it is a_1 in the numerator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{2k}=\frac{a_0}{(2k)!}\] \a_{2k+1}=\frac{a_1}{2k+1}[\] I didn't forgot...I just checking if you were paying attention :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{2k+1}=\frac{a_1}{2k+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{2k+1}=\frac{a_1}{(2k+1)!}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait there is more...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1Much 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.

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you've got the right idea, but we need to use superposition to get the full answer

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0put them into one formula somehow

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh y and c_1 and so on and so forth...hold on

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0gosh the sums probably shouldn't be there...

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can I put my laundry in the dryer?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1nah, 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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so 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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1awesome! 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sounds 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...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will. Thanks! quick question, did you make your own empanadas?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1nope, bought 'em at the corner I only cook hamburgers and pasta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what? you've been in Mexica for a little while now! Gotta learn to cook like the locals!

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1well I make quesadillas, I guess that's a start lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep, work your way up from here :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348252365896:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348252603116:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348252694415:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1348252778070:dw
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