## xEnOnn Group Title 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 $r_2$? 2 years ago 2 years ago

1. dichiaraj Group Title

is your summation suppose to have limits?

2. xEnOnn Group Title

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

3. dichiaraj Group Title

true

4. dichiaraj Group Title

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

5. xEnOnn Group Title

why?

6. dichiaraj Group Title

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 Group Title

havent*

8. xEnOnn Group Title

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 Group Title

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 Group Title

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

11. dichiaraj Group Title

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

12. xEnOnn Group Title

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 Group Title

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

14. xEnOnn Group Title

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

15. dumbcow Group Title

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

16. xEnOnn Group Title

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 Group Title

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 Group Title

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

19. dichiaraj Group Title

its ok i would be doing the same thing =)

20. dumbcow Group Title

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 Group Title

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 Group Title

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 Group Title

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

24. dichiaraj Group Title

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 Group Title

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 Group Title

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

27. dumbcow Group Title

i think this is it

28. dichiaraj Group Title

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 Group Title

what do you mean by unknowns?

30. xEnOnn Group Title

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

31. dumbcow Group Title

oh

32. dichiaraj Group Title

yea thats what i meant

33. dumbcow Group Title

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

34. dumbcow Group Title

however the correct way of writing that is in summation form

35. dumbcow Group Title

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

36. dichiaraj Group Title

your solving for r1 and r2 i believe

37. dichiaraj Group Title

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

38. dumbcow Group Title

oh got it thanks

39. dumbcow Group Title

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