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 \[r_2\]?

- anonymous

- jamiebookeater

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- anonymous

is your summation suppose to have limits?

- anonymous

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

- anonymous

true

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## More answers

- anonymous

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

- anonymous

why?

- 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
]

- anonymous

havent*

- 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...

- 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

- anonymous

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

- anonymous

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

- 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...

- dumbcow

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

- anonymous

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

- dumbcow

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

- 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...

- 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

- anonymous

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

- anonymous

its ok i would be doing the same thing =)

- dumbcow

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

- 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 =)

- 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.

- dumbcow

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

- 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

- 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.

- dumbcow

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

- dumbcow

i think this is it

- 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

- dumbcow

what do you mean by unknowns?

- anonymous

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

- dumbcow

oh

- anonymous

yea thats what i meant

- dumbcow

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

- dumbcow

however the correct way of writing that is in summation form

- dumbcow

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

- anonymous

your solving for r1 and r2 i believe

- anonymous

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

- dumbcow

oh got it
thanks

- dumbcow

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

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