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xEnOnn
From this: \[(1+t+t^2)^5=\sum \binom{5}{r_1}\binom{5-r_1}{r_2}t^{r_2}t^{2r_1}\] How do I carry on to find out the values of \[r_1\] and \[r_2\]?
is your summation suppose to have limits?
yea...but this is just binomial so i just lazy to put them in...
that has to be the ugliest summation question i have ever seen
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 ]
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 co-effiicient value and then put them into an equation to solve for the r... but i'm stuck at finding the co-efficient value...
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
i'm actually on wikipedia now... but i just couldn't figure out... i'm still trying though...
Ha ha I'm on it too trying to remember this stuff. lols
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...
oh i see what you're doing just trying to expand yeah its not a binomial, its a trinomial
I think I need the values of \[r_1\] and \[r_2\] to expand, wouldn't I?
not sure with a binomial, they would sum up to 5
hmm... could it be \[r_2=5-r_1\] 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...
your first equation i believe is wrong, r_2=5-r_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
yea you are right. that is wrong. i was trying different ways to figure out an answer to it. haha..
its ok i would be doing the same thing =)
can't quite get the summation to work out here's the expansion though \[= t^{10}+5t^{9}+15t^{8}+30t^{7}+44t^{6}+50t^{5}+44t^{4}+30t^{3}+15t^{2}+5t+1\]
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 =)
Would this be right if I say \[r_2+2r_1=5\] and that \[r_2=5-r_1\] and then solve simultaneously, to get \[r_1=1,r_2=5/2\]? But I remember that can't be a fraction.
doesn't seem right r1+r2 = 5 2r1 +r2 = 5 -> r1 = 2r1 -> 1=2 Equations contradict each other
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
The one I got was because I thought from \[(1+t+t^2)^5=\sum \binom{5}{r_1}\binom{5-r_1}{r_2}t^{r_2}t^{2r_1}\] The degree of \[r_2\] was created out of \[5-r_1\] and \[2r_1\] was created out of the 5 in the \[ \binom{5}{r_1}\]. But I'm not sure if this is right.
\[\sum_{k=0}^{5}\sum_{j=0}^{5-k}\sum_{i=0}^{5-j-k}\frac{5!}{i!j!k!}t^{j}t^{2k}\]
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
what do you mean by unknowns?
I think he mean the k and j are unknowns and are needed to find the \[r_1\] and \[r_2\].
yea thats what i meant
well you are looking at every possible combination of i,j,k where sum equals 5
however the correct way of writing that is in summation form
sorry im kinda lost on how r1 and r2 have anything to do with it
your solving for r1 and r2 i believe
in word form i believe the question would read what values for r1 and r2 would make this equation true
you said the 5-r1 is incorrect, what should it be?