Question

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 \[r_2\]?

Answer 1

is your summation suppose to have limits?

Answer 2

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

Answer 3

true

Answer 4

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

Answer 5

why?

Answer 6

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
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Answer 7

havent*

Answer 8

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...

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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...

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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...

Answer 17

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

Answer 18

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

Answer 19

its ok i would be doing the same thing =)

Answer 20

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

Answer 21

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 =)

Answer 22

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.

Answer 23

doesn't seem right
r1+r2 = 5
2r1 +r2 = 5
-> r1 = 2r1
-> 1=2

Equations contradict each other

Answer 24

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

Answer 25

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.

Answer 26

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

Answer 27

i think this is it

Answer 28

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

Answer 29

what do you mean by unknowns?

Answer 30

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

Answer 31

oh

Answer 32

yea thats what i meant

Answer 33

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

Answer 34

however the correct way of writing that is in summation form

Answer 35

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

Answer 36

your solving for r1 and r2 i believe

Answer 37

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

Answer 38

oh got it
thanks

Answer 39

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