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xEnOnn

  • 3 years ago

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 \[r_2\]?

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  1. dichiaraj
    • 3 years ago
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    is your summation suppose to have limits?

  2. xEnOnn
    • 3 years ago
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    yea...but this is just binomial so i just lazy to put them in...

  3. dichiaraj
    • 3 years ago
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    true

  4. dichiaraj
    • 3 years ago
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    that has to be the ugliest summation question i have ever seen

  5. xEnOnn
    • 3 years ago
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    why?

  6. dichiaraj
    • 3 years ago
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    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 ]

  7. dichiaraj
    • 3 years ago
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    havent*

  8. xEnOnn
    • 3 years ago
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    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...

  9. dichiaraj
    • 3 years ago
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    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

  10. xEnOnn
    • 3 years ago
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    i'm actually on wikipedia now... but i just couldn't figure out... i'm still trying though...

  11. dichiaraj
    • 3 years ago
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    Ha ha I'm on it too trying to remember this stuff. lols

  12. xEnOnn
    • 3 years ago
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    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...

  13. dumbcow
    • 3 years ago
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    oh i see what you're doing just trying to expand yeah its not a binomial, its a trinomial

  14. xEnOnn
    • 3 years ago
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    I think I need the values of \[r_1\] and \[r_2\] to expand, wouldn't I?

  15. dumbcow
    • 3 years ago
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    not sure with a binomial, they would sum up to 5

  16. xEnOnn
    • 3 years ago
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    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...

  17. dichiaraj
    • 3 years ago
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    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

  18. xEnOnn
    • 3 years ago
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    yea you are right. that is wrong. i was trying different ways to figure out an answer to it. haha..

  19. dichiaraj
    • 3 years ago
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    its ok i would be doing the same thing =)

  20. dumbcow
    • 3 years ago
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    can't quite get the summation to work out here's the expansion though \[= t^{10}+5t^{9}+15t^{8}+30t^{7}+44t^{6}+50t^{5}+44t^{4}+30t^{3}+15t^{2}+5t+1\]

  21. dichiaraj
    • 3 years ago
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    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 =)

  22. xEnOnn
    • 3 years ago
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    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.

  23. dumbcow
    • 3 years ago
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    doesn't seem right r1+r2 = 5 2r1 +r2 = 5 -> r1 = 2r1 -> 1=2 Equations contradict each other

  24. dichiaraj
    • 3 years ago
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    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

  25. xEnOnn
    • 3 years ago
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    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.

  26. dumbcow
    • 3 years ago
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    \[\sum_{k=0}^{5}\sum_{j=0}^{5-k}\sum_{i=0}^{5-j-k}\frac{5!}{i!j!k!}t^{j}t^{2k}\]

  27. dumbcow
    • 3 years ago
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    i think this is it

  28. dichiaraj
    • 3 years ago
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    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

  29. dumbcow
    • 3 years ago
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    what do you mean by unknowns?

  30. xEnOnn
    • 3 years ago
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    I think he mean the k and j are unknowns and are needed to find the \[r_1\] and \[r_2\].

  31. dumbcow
    • 3 years ago
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    oh

  32. dichiaraj
    • 3 years ago
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    yea thats what i meant

  33. dumbcow
    • 3 years ago
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    well you are looking at every possible combination of i,j,k where sum equals 5

  34. dumbcow
    • 3 years ago
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    however the correct way of writing that is in summation form

  35. dumbcow
    • 3 years ago
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    sorry im kinda lost on how r1 and r2 have anything to do with it

  36. dichiaraj
    • 3 years ago
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    your solving for r1 and r2 i believe

  37. dichiaraj
    • 3 years ago
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    in word form i believe the question would read what values for r1 and r2 would make this equation true

  38. dumbcow
    • 3 years ago
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    oh got it thanks

  39. dumbcow
    • 3 years ago
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    you said the 5-r1 is incorrect, what should it be?

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