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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 \[r_2\]?
 2 years ago
 2 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 \[r_2\]?
 2 years ago
 2 years ago

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dichiarajBest ResponseYou've already chosen the best response.2
is your summation suppose to have limits?
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
that has to be the ugliest summation question i have ever seen
 2 years ago

dichiarajBest ResponseYou've already chosen the best response.2
just 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 ]
 2 years ago

xEnOnnBest ResponseYou've already chosen the best response.0
i 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...
 2 years ago

dichiarajBest ResponseYou've already chosen the best response.2
I 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
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
Ha ha I'm on it too trying to remember this stuff. lols
 2 years ago

xEnOnnBest ResponseYou've already chosen the best response.0
lol...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...
 2 years ago

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

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

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

xEnOnnBest ResponseYou've already chosen the best response.0
hmm... 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...
 2 years ago

dichiarajBest ResponseYou've already chosen the best response.2
your 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
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
its ok i would be doing the same thing =)
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
Well 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 =)
 2 years ago

xEnOnnBest ResponseYou've already chosen the best response.0
Would 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.
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
where 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
 2 years ago

xEnOnnBest ResponseYou've already chosen the best response.0
The 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.
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
yes 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
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
what do you mean by unknowns?
 2 years ago

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

dichiarajBest ResponseYou've already chosen the best response.2
yea thats what i meant
 2 years ago

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

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

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

dichiarajBest ResponseYou've already chosen the best response.2
your solving for r1 and r2 i believe
 2 years ago

dichiarajBest ResponseYou've already chosen the best response.2
in word form i believe the question would read what values for r1 and r2 would make this equation true
 2 years ago

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