Binomial Theorem help? (1 Viewer)

[shinji]

hey. im kinda stuck on a question ... and yerh.

here's the question:

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 of 8:27

would be helpful if u stated the correct answer and how u did it. thanks!

 

[Mountain.Dew]

hey. im kinda stuck on a question ... and yerh.

here's the question:

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 of 8:27

would be helpful if u stated the correct answer and how u did it. thanks!

heres more of my 2 cents.

simply put, this question translates to be ==> ⁿC_r : ⁿC_n-r = 8 : 27

or simply, ⁿC_r/ⁿC_n-r = 8/27, where n = 15.

now, finding the 'r'th term ==> T_r = ⁿC_r-1a^n-(r-1)b^r-1 = ¹⁵C_r-1(2x)^15-(r-1)3^r-1 =
¹⁵C_r-1(2)^15-(r-1)3^r-1(x)^15-(r-1) =

so, coefficient of T_r = ¹⁵C_r-1(2)^15-(r-1)3^r-1

...more to come soon.

 

[clue]

:wave: the correct answer is r=9

heres my solution
T(r)=15Cr * (2x)^r * 3^(15-r)

T(15-r)= 15C(15-r) * (2x)^(15-r) * 3^r

T(r)/T(15-r)= (2x)^(2r-15) * 3^(15-2r) (nCk=nC(n-k))

therefore

2^(2r-15) * 3^(15-2r)=8/27

(3/2)^15 * (2/3)^2r=8/27

(2/3)^2r= (2/3)^3 * (2/3)^15
= (2/3)^18

taking the log of both sides should give you
2r=18 r=9

i hope that makes sense, as i did miss a few steps here and there, feel free to ask for further explanation

btw i hope by maths notation is right-this is the first time i've written maths solutions on a computer...

 

[clue]

mountian dew how did you manage to write the powers?
thanks in advance for an answer...

 

[Mountain.Dew]

mountian dew how did you manage to write the powers?
thanks in advance for an answer...

ah very simple. say you wanted x cubed, or x^3. you want "x³", yes?

simply type in "x(sup)<insert the value of the power>(/sup)" BUT replace the ( ) with [ ]

 

[shinji]

:wave: the correct answer is r=9

heres my solution
T(r)=15Cr * (2x)^r * 3^(15-r)

T(15-r)= 15C(15-r) * (2x)^(15-r) * 3^r

T(r)/T(15-r)= (2x)^(2r-15) * 3^(15-2r) (nCk=nC(n-k))

therefore

2^(2r-15) * 3^(15-2r)=8/27

(3/2)^15 * (2/3)^2r=8/27

(2/3)^2r= (2/3)^3 * (2/3)^15
= (2/3)^18

taking the log of both sides should give you
2r=18 r=9

i hope that makes sense, as i did miss a few steps here and there, feel free to ask for further explanation

btw i hope by maths notation is right-this is the first time i've written maths solutions on a computer...

lol thanks for the effort. but i don't really understand this line

T(r)/T(15-r)= (2x)^(2r-15) * 3^(15-2r) (nCk=nC(n-k))

 

[shinji]

erm. help? ''

=/

 

[shinji]

oh yeah btw;

i looked at the answers; the rTh term is 7.

=/; ne way on how to find that out?

 

[shinji]

hey. finally got the answer for those who are interested.

(2x+3)¹⁵

Begin Tr^th : ¹⁵C_r-1 (2x)^16-r 3^r-1

End Tr^th : ¹⁵C_16-r (2x)^r-1 3^16-r

Tr/End Tr = 8 / 27 = 15!/(r-1)!(16-r)! * 2^16-r/2^r-1 * 3^r-1/3^16-r * (16-r)!(r-1)!/15!

8/27 = 2^17-2r/3^17-2r

2³/3³ = 2^17-2r/3^17-2r

.:. 17-2r = 3
-2r = -14
r = 7

which is the correct answer ~_~

 

[Mountain.Dew]

YAY! congrats shinji!

 

[shinji]

haha thanks m.d ! =D