## anonymous 5 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 $r_2$?

1. anonymous

is your summation suppose to have limits?

2. anonymous

yea...but this is just binomial so i just lazy to put them in...

3. anonymous

true

4. anonymous

that has to be the ugliest summation question i have ever seen

5. anonymous

why?

6. anonymous

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. anonymous

havent*

8. anonymous

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. anonymous

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. anonymous

i'm actually on wikipedia now... but i just couldn't figure out... i'm still trying though...

11. anonymous

Ha ha I'm on it too trying to remember this stuff. lols

12. anonymous

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. anonymous

oh i see what you're doing just trying to expand yeah its not a binomial, its a trinomial

14. anonymous

I think I need the values of $r_1$ and $r_2$ to expand, wouldn't I?

15. anonymous

not sure with a binomial, they would sum up to 5

16. anonymous

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. anonymous

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. anonymous

yea you are right. that is wrong. i was trying different ways to figure out an answer to it. haha..

19. anonymous

its ok i would be doing the same thing =)

20. anonymous

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. anonymous

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. anonymous

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. anonymous

doesn't seem right r1+r2 = 5 2r1 +r2 = 5 -> r1 = 2r1 -> 1=2 Equations contradict each other

24. anonymous

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. anonymous

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. anonymous

$\sum_{k=0}^{5}\sum_{j=0}^{5-k}\sum_{i=0}^{5-j-k}\frac{5!}{i!j!k!}t^{j}t^{2k}$

27. anonymous

i think this is it

28. anonymous

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. anonymous

what do you mean by unknowns?

30. anonymous

I think he mean the k and j are unknowns and are needed to find the $r_1$ and $r_2$.

31. anonymous

oh

32. anonymous

yea thats what i meant

33. anonymous

well you are looking at every possible combination of i,j,k where sum equals 5

34. anonymous

however the correct way of writing that is in summation form

35. anonymous

sorry im kinda lost on how r1 and r2 have anything to do with it

36. anonymous

your solving for r1 and r2 i believe

37. anonymous

in word form i believe the question would read what values for r1 and r2 would make this equation true

38. anonymous

oh got it thanks

39. anonymous

you said the 5-r1 is incorrect, what should it be?