Binomial Theorem Q (2 Viewers)

[lyounamu]

Fitzpatrick 3 Unit textbook Q40, page 209:

Find the value of r if the coefficients of the rth term from the beginning and the rth term from the end of (2x+3)^15 are in the ratio 8:27.

I think I approached this question correctly (using the simulatenous in the end but cannot get the right answer).

I just want to know how to approach this question correctly (because mine didn't work).

Thanks!

 

[conics2008]

Hey.

(2x+3)^15 The expansion is 15CR right ???

15cr*(2x)^15*(3)^0 15c(r+1)(2x)^15-r*(3)^r .... try picking any two and then dividing them and make them equal to 8/27


how does this approach sound!!

sorry but its been almost a year i havn't touched bionimal theorem.

 

[lyounamu]

Hey.

(2x+3)^15 The expansion is 15CR right ???

15cr*(2x)^15*(3)^0 15c(r+1)(2x)^15-r*(3)^r .... try picking any two and then dividing them and make them equal to 8/27


how does this approach sound!!

sorry but its been almost a year i havn't touched bionimal theorem.

No. That wouldn't work.

What I did was:

The rth term from the beginning: 15Cr-1 . (2x)^(16-r) . 3^(r-1)
The rth term from the end: 15C15-r . (2x)^r . 3^(15-r)

The coefficient of the rth term from the beginning: 15Cr-1 . 2^(16-r) . 3^(r-1) = 8 ...(1)
The coefficient of the rth term from the end: 15C15-r . 2^r . 3^(15-r) = 27 ...(2)

And kinda solve it from there. But I get really weird (massive) answer at the end.

 

[conics2008]

wait ill get my text book.. can you give me the answer please.

 

[lyounamu]

wait ill get my text book.. can you give me the answer please.

The answer is 7.

 

[conics2008]

sorry buddy, my memory of using binomial theorem has gone..

i looked back, but coudn't get it through my head.. looks like i have to study this all over again.. I hope someone could come and help you out..

 

[lyounamu]

sorry buddy, my memory of using binomial theorem has gone..

i looked back, but coudn't get it through my head.. looks like i have to study this all over again.. I hope someone could come and help you out..

It's okay, mate. That happens all the time for me. (I forget occasionally)

 

[3unitz]

Fitzpatrick 3 Unit textbook Q40, page 209:

Find the value of r if the coefficients of the rth term from the beginning and the rth term from the end of (2x+3)^15 are in the ratio 8:27.

I think I approached this question correctly (using the simulatenous in the end but cannot get the right answer).

I just want to know how to approach this question correctly (because mine didn't work).

Thanks!

(2x + 3)^15

T_rb = 15C(r-1) 2^(16-r) 3^(r-1) (from beginning)
T_re = 15C(16-r) 2^(r-1) 3^(16-r) (from end)

T_rb / T_re = 8/27
8/27 = 15C(r-1) 2^(16-r) 3^(r-1) / 15C(16-r) 2^(r-1) 3^(16-r)
= [15C(r-1) / 15C(16-r)] . [2^(16-r) / 2^(r-1)] . [3^(r-1) / 3^(16-r)]
= {15!/[(r-1)!(16-r)!] / 15!/(16-r)!(r-1)!} . 2^(16-r - r +1) . 3^(r - 1 - 16 + r)
= 2^(17 - 2r) . 3^(2r - 17)

ln(8/27) = (17 - 2r)ln2 + (2r - 17)ln3
ln(8/27) = 17ln2 - 2rln2 + 2rln3 - 17ln3
ln(8/27) + 17ln3 - 17ln2 = r(2ln3 - 2ln2)
r = [ln(8/27) + 17ln3 - 17ln2] / (2ln3 - 2ln2)
r = 7

 

[lolokay]

you know that the rth term from beginning and end will have the same combiantions part of their coefficient

and you can see that 8 = 2^3 and 27 = 3^3, so rth term from the end must be 3 places higher than the rth term, and since they each have the same number of places, x, on the other side of them, and there are a total of 16 terms
you have x + 1 + 2 + 1 + x = 16, so x = 6. r is the place after x, so r = 7

 

[lyounamu]

(2x + 3)^15

T_rb = 15C(r-1) 2^(16-r) 3^(r-1) (from beginning)
T_re = 15C(16-r) 2^(r-1) 3^(16-r) (from end)

T_rb / T_re = 8/27
8/27 = 15C(r-1) 2^(16-r) 3^(r-1) / 15C(16-r) 2^(r-1) 3^(16-r)
= [15C(r-1) / 15C(16-r)] . [2^(16-r) / 2^(r-1)] . [3^(r-1) / 3^(16-r)]
= {15!/[(r-1)!(16-r)!] / 15!/(16-r)!(r-1)!} . 2^(16-r - r +1) . 3^(r - 1 - 16 + r)
= 2^(17 - 2r) . 3^(2r - 17)

ln(8/27) = (17 - 2r)ln2 + (2r - 17)ln3
ln(8/27) = 17ln2 - 2rln2 + 2rln3 - 17ln3
ln(8/27) + 17ln3 - 17ln2 = r(2ln3 - 2ln2)
r = [ln(8/27) + 17ln3 - 17ln2] / (2ln3 - 2ln2)
r = 7

Thanks!

 

[lyounamu]

you know that the rth term from beginning and end will have the same combiantions part of their coefficient

and you can see that 8 = 2^3 and 27 = 3^3, so rth term from the end must be 3 places higher than the rth term, and since they each have the same number of places, x, on the other side of them, and there are a total of 16 terms
you have x + 1 + 2 + 1 + x = 16, so x = 6. r is the place after x, so r = 7

I knew that but I just couldn't write that as my answer, lol. Thanks for your great insight by the way!

 

[lolokay]

I knew that but I just couldn't write that as my answer, lol. Thanks for your great insight by the way!

well you could express that mathematically,

( 15Cr*2^16-r * 3^r-1 )/8 = (15C(15-r)*2^r-1 * 3^16-r)/27
2^16-r * 3^r-1 * 3^3 = 2^r-1 * 3^16-r * 2^3
16 - r = r-1 + 3
2r = 14
r = 7

 

[lyounamu]

well you could express that mathematically,

( 15Cr*2^16-r * 3^r-1 )/8 = (15C(15-r)*2^r-1 * 3^16-r)/27
2^16-r * 3^r-1 * 3^3 = 2^r-1 * 3^16-r * 2^3
16 - r = r-1 + 3
2r = 14
r = 7

Yeah, I get it. Thanks. :uhhuh: