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anonymous
 5 years ago
From this: \[(1+t+t^2)^5=\sum \binom{5}{r_1}\binom{5r_1}{r_2}t^{r_2}t^{2r_1}\]
How do I carry on to find out the values of \[r_1\] and \[r_2\]?
anonymous
 5 years ago
From this: \[(1+t+t^2)^5=\sum \binom{5}{r_1}\binom{5r_1}{r_2}t^{r_2}t^{2r_1}\] How do I carry on to find out the values of \[r_1\] and \[r_2\]?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is your summation suppose to have limits?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea...but this is just binomial so i just lazy to put them in...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that has to be the ugliest summation question i have ever seen

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just looks ugly i have worked with these in like a couple years, so i guess to me its just looks crazy since my field doesn't use frequently; meaning i have fallen out of practice. ha ha ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have done this quite some time ago like last yr..but i recently need these again but forgotten how to continue from here...i remember i need to find out the coeffiicient value and then put them into an equation to solve for the r... but i'm stuck at finding the coefficient value...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would go to wikipedia and search binomial theorem and see if you can glean the answer from that. It might help with the expansion of the binomial coefficient

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm actually on wikipedia now... but i just couldn't figure out... i'm still trying though...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ha ha I'm on it too trying to remember this stuff. lols

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol...the idea was to expand that \[(1+t+t^2)^5\] that got me started using binomial...and I thought I could do it until I tried that I realise i couldn't...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see what you're doing just trying to expand yeah its not a binomial, its a trinomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I need the values of \[r_1\] and \[r_2\] to expand, wouldn't I?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure with a binomial, they would sum up to 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm... could it be \[r_2=5r_1\] and \[2r_1=5\]? But this would cause \[r_1=5/2\] which can't happen in such a thing since the coefficients must be integers... hmm...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your first equation i believe is wrong, r_2=5r_1 is incorrect. If that were true one your binomial coefficients would be 1 and that seems like unlikely simplification, but i could be wrong thats just my input

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea you are right. that is wrong. i was trying different ways to figure out an answer to it. haha..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its ok i would be doing the same thing =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can't quite get the summation to work out here's the expansion though \[= t^{10}+5t^{9}+15t^{8}+30t^{7}+44t^{6}+50t^{5}+44t^{4}+30t^{3}+15t^{2}+5t+1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well i think i got something that might help you. Knowing that this is a trinomial it must follow the form \[(a+b+c)^n=\sum_{i,j,k}^{n}\left(\begin{matrix}n \\ i,j,k\end{matrix}\right)*a^I*b^j*c^k\] meaning: \[t^{r_2}+t^{2r_1}=t^5\] since: i+j+k=5 which implies that \[r_2+2r_1=5\] thats as far as I've gotten and I'm tired hope maybe this inspires someone else to finish =)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Would this be right if I say \[r_2+2r_1=5\] and that \[r_2=5r_1\] and then solve simultaneously, to get \[r_1=1,r_2=5/2\]? But I remember that can't be a fraction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0doesn't seem right r1+r2 = 5 2r1 +r2 = 5 > r1 = 2r1 > 1=2 Equations contradict each other

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get r1+r2=5 from... I'm pretty sure one of the necessary equations is 2r1+r2=5. k=2r1 always, j=r2, and i=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The one I got was because I thought from \[(1+t+t^2)^5=\sum \binom{5}{r_1}\binom{5r_1}{r_2}t^{r_2}t^{2r_1}\] The degree of \[r_2\] was created out of \[5r_1\] and \[2r_1\] was created out of the 5 in the \[ \binom{5}{r_1}\]. But I'm not sure if this is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=0}^{5}\sum_{j=0}^{5k}\sum_{i=0}^{5jk}\frac{5!}{i!j!k!}t^{j}t^{2k}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i agree that is a correct way of writing it, but even knowing that i=0 you still have two unknowns and only one equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean by unknowns?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think he mean the k and j are unknowns and are needed to find the \[r_1\] and \[r_2\].

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea thats what i meant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you are looking at every possible combination of i,j,k where sum equals 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0however the correct way of writing that is in summation form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry im kinda lost on how r1 and r2 have anything to do with it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your solving for r1 and r2 i believe

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in word form i believe the question would read what values for r1 and r2 would make this equation true

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you said the 5r1 is incorrect, what should it be?
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