# [SOLVED] binomial theorem =(

• February 17th 2009, 01:03 AM
teddybear67
[SOLVED] binomial theorem =(
hey, ive got a question here in which ive been pondering over for hours, hope someone could help me ^^

If the coefficients of $x^k$ and $x^k+1$ in the expansion of $(2 + 3x)^(10)$ are equal, find K.

• February 17th 2009, 01:48 AM
Quote:

Originally Posted by teddybear67
hey, ive got a question here in which ive been pondering over for hours, hope someone could help me ^^

If the coefficients of $x^k$ and $x^{k+1}$ in the expansion of $(2 + 3x)^{10}$ are equal, find K.

$T_n$ is the coefficient of nth term
$
T_k= T_{k+1}
$

$
^nC_k 2^k 3^{n-k} = ~^nC_{k+1} 2^{k+1} 3^{n-k-1}
$

Put
n= 10
And find k
• February 17th 2009, 03:18 AM
teddybear67
yea, thats where i got lost.
btw sorry the 10 is supposed to be 19, so its
$
(2 + 3x)^{19}
$

i dont think it would make any difference in the working u gave me anyways. lols
anywyas, my working is
$
^{19}C_k 2^k 3^{n-k} = ^{19}C_{k+1} 2^{k+1} 3^{n-k-1}$

=
$\frac{^{19}C_k} {^{19}C_{k+1}} = 2 X 3$

$\frac{^{19}C_k} {^{19}C_{k+1}} = 6$

and i got stuck here...
• February 17th 2009, 03:54 AM
Plato
${19 \choose 9} = {19 \choose 10}$
• February 17th 2009, 03:58 AM
teddybear67
i dont get it...
i know what ${19 \choose 9} = {19 \choose 10}$ is
buthow did u get 9 and 10?
• February 17th 2009, 04:15 AM
don mean to post .
• February 17th 2009, 05:07 AM
Binomial Coefficients
Hello teddybear67
Quote:

Originally Posted by teddybear67
yea, thats where i got lost.
btw sorry the 10 is supposed to be 19, so its
$
(2 + 3x)^{10}
$

i dont think it would make any difference in the working u gave me anyways. lols
anywyas, my working is
$
^{19}C_k 2^k 3^{n-k} = ^{19}C_{k+1} 2^{k+1} 3^{n-k-1}$

=
$\frac{^{19}C_k} {^{19}C_{k+1}} = \color{red}{2 X 3}$

$\frac{^{19}C_k} {^{19}C_{k+1}} = 6$

and i got stuck here...

$\frac{^{19}C_k} {^{19}C_{k+1}} =\frac{2}{3}$

$\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}$

$\Rightarrow \frac{k+1}{19-k}=\frac{2}{3}$, if you cancel carefully

$\Rightarrow 3k+3 = 38 - 2k$

$\Rightarrow k = 7$

• February 17th 2009, 05:37 AM
teddybear67
thanks alot grandad, but i dont get this part :
$
\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}
$

sorry, im confused over the factorial and the "choose". its the 1 part of binomial theorem in which i never get ><

oh ya, and the answer at the back of the book in which i got the question from is 11 and not 7. Do you know why? or is the book wrong? because there are times in which the book gives wrong answers...
• February 17th 2009, 05:47 AM
Quote:

Originally Posted by teddybear67
thanks alot grandad, but i dont get this part :
$
\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}
$

sorry, im confused over the factorial and the "choose". its the 1 part of binomial theorem in which i never get ><

oh ya, and the answer at the back of the book in which i got the question from is 11 and not 7. Do you know why? or is the book wrong? because there are times in which the book gives wrong answers...

Use that fact that ${n \choose r}=\frac{n!}{(n-r)!r!}$
• February 17th 2009, 05:59 AM
Binomial Coefficients
Hello teddybear67
Quote:

Originally Posted by teddybear67
thanks alot grandad, but i dont get this part :
$
\Rightarrow \frac{19!(k+1)!(19-[k+1])!}{k!(19-k)!19!} = \frac{2}{3}
$

sorry, im confused over the factorial and the "choose". its the 1 part of binomial theorem in which i never get ><

oh ya, and the answer at the back of the book in which i got the question from is 11 and not 7. Do you know why? or is the book wrong? because there are times in which the book gives wrong answers...

Yes, it does look rather a mess, doesn't it?

The working goes like this:

$\frac{^{19}C_k}{^{19}C_{k+1}}=\frac{2}{3}$ (I assume you agree that this is correct!)

$\Rightarrow \frac{19!}{k!(19-k)!}\div \frac{19!}{(k+1)!(19-[k+1]!)} =\frac{2}{3}$

$\Rightarrow \frac{19!}{k!(19-k)!}\times \frac{(k+1)!(19-[k+1]!)}{19!} =\frac{2}{3}$

Now the $19!$ on the top and bottom cancel immediately. Then notice that the $(k+1)!$ term on the top = $(k+1) \times k!$. So it cancels with the $k!$ term on the bottom to leave $(k+1)$ on the top.

Similarly the $(19-k)!$ term on the bottom is $(19-k) \times (19-[k+1])!$. So it cancels with the term on the top to leave $(19-k)$ on the bottom.

Hence $\frac{k+1}{19-k}=\frac{2}{3}$

... etc

I have checked my answer using an Excel spreadsheet (see the attached) and it is correct! So the answer in the book is wrong (nothing new there!).

• February 17th 2009, 06:05 AM
teddybear67
lol i assumed that your
$
\frac{^{19}C_k}{^{19}C_{k+1}}=\frac{2}{3}
$

is correct, but i double checked and its supposed to be 3/2 . sorry!!

but yea, i understand your workings and managed to get the hang of it. thanks alot grandad!!! ^^
• February 17th 2009, 06:16 AM
Binomial Coefficients
Hello teddybear67
Quote:

Originally Posted by teddybear67
lol i assumed that your
$
\frac{^{19}C_k}{^{19}C_{k+1}}=\frac{2}{3}
$

is correct, but i double checked and its supposed to be 3/2 . sorry!!

but yea, i understand your workings and managed to get the hang of it. thanks alot grandad!!! ^^

Ah, I see where $k = 11$ comes from now. Of course, it depends which way round you're expanding the binomial. Your original question has the expression as $(2 + 3x)^{19}$. And for this the answer is $k = 7$, and the equal coefficients are the 7th and 8th.

But if you expand it the other way round $(3x+2)^{19}$, then the equal coefficients are the (19 - 7)th and (19 - 8)th i.e. the 12 th and 11th.

It's all so simple really.