Thread: Binomial coefficient in an expansion

1. Binomial coefficient in an expansion

Hello.

I must "use the binomial theorem to determine the coefficient x^14 in the expansion of (3x^2-1/3)^16".

I find that:
n=16
r=14
a=3x^2 (I believe)
b=-1/3

I write that (sorry but the matrix brackets I cannot make):

(16) (3x^2) ^(16-14+1) (-1/3)^13
(13)

= (16) (3x^2)^3 (-1/3)^13
(13)

= (16!/(16-13)!13!) (1/? x^6)

This part is how I am getting confused.

Can somebody help me please to continue to solving this question? I must "expressing the coefficient as a fraction in lowest terms" but I don't think I have calculated well above to make the fraction to continue. The book is not showing this kind of difficult examples before we must do the assignment!

Thank you.
Brigitte

2. Originally Posted by dolkam
Hello.

I must "use the binomial theorem to determine the coefficient x^14 in the expansion of (3x^2-1/3)^16".

I find that:
n=16
r=14
a=3x^2 (I believe)
b=-1/3

I write that (sorry but the matrix brackets I cannot make):

(16) (3x^2) ^(16-14+1) (-1/3)^13
(13)

= (16) (3x^2)^3 (-1/3)^13
(13)

= (16!/(16-13)!13!) (1/? x^6)

This part is how I am getting confused.

Can somebody help me please to continue to solving this question? I must "expressing the coefficient as a fraction in lowest terms" but I don't think I have calculated well above to make the fraction to continue. The book is not showing this kind of difficult examples before we must do the assignment!

Thank you.
Brigitte
$
(x+y)^n = {n \choose 0}x^n y^0 + {n \choose 1}x^{n-1}y^1 + {n \choose 2}x^{n-2}y^2 + {n \choose 3}x^{n-3}y^3 + ... + {n \choose n}y^n$

In your case $x = 3x^2$ and $y= -\frac{1}{3}$

Since $-\frac{1}{3}$ does not contain an x term then raising it to the power so the power is unaffected.

As we have an $x^2$ term the coefficient of x^14 will be given when n=7. Use the binomial theorem for the 7th term to find the coefficient

Spoiler:
From the binomial theorem ${16 \choose 7}(x^2)^7 \cdot \left(-\frac{1}{3}\right)^{16-7}$

Hence the coefficient is ${16 \choose 7} \cdot \left(-\frac{1}{3}\right)^{9}$.

Be aware that ${16 \choose 7} = \frac{16!}{9!7!}$ whereas $\left(-\frac{1}{3}\right)^{9}$ is simply minus one-third raised to the 9th power

3. Originally Posted by dolkam
Hello.

I must "use the binomial theorem to determine the coefficient x^14 in the expansion of (3x^2-1/3)^16".

I find that:
n=16
r=14
a=3x^2 (I believe)
b=-1/3

I write that (sorry but the matrix brackets I cannot make):

(16) (3x^2) ^(16-14+1) (-1/3)^13
(13)

= (16) (3x^2)^3 (-1/3)^13
(13)

= (16!/(16-13)!13!) (1/? x^6)

This part is how I am getting confused.

Can somebody help me please to continue to solving this question? I must "expressing the coefficient as a fraction in lowest terms" but I don't think I have calculated well above to make the fraction to continue. The book is not showing this kind of difficult examples before we must do the assignment!

Thank you.
Brigitte
The general term is ${16 \choose r} (3x^2)^{16 - r} \left(-\frac{1}{3}\right)^r = {16 \choose r} (3x^2)^{16 - r} (-1)^r (3)^{-r}$ $= {16 \choose r} (-1)^r 3^{16-r} 3^{-r} (x^2)^{(16 - r)} = {16 \choose r} (-1)^r 3^{16 - 2r} x^{2(16-r)}$.

You require the value of the coefficient ${16 \choose r} (-1)^r 3^{16 - 2r}$ when $2(16-r) = 14$.

4. I am a little confused

Tthank you for the replies.

I am following all of the details of your work but I am not understanding how you can be making a different formula to my formula.

I have the binomial formula from class (sorry I cannot make the brackets):

(n) * a^(n-r+1) * b^(r-1)
(r-1)

(n) * a^(n-r) * b^r
(r)

Please is this formulas the same things?

Perhaps this is the reason why I am not making the same answer when I am working the problem?

I will appreciate it if you can be telling how this is the same.

Thanks.
Brigitte

5. Originally Posted by dolkam
Tthank you for the replies.

I am following all of the details of your work but I am not understanding how you can be making a different formula to my formula.

I have the binomial formula from class (sorry I cannot make the brackets):

(n) * a^(n-r+1) * b^(r-1)
(r-1)

(n) * a^(n-r) * b^r
(r)

Please is this formulas the same things?

Perhaps this is the reason why I am not making the same answer when I am working the problem?

I will appreciate it if you can be telling how this is the same.

Thanks.
Brigitte
Both formulas are equivalent. If you prefer using yours to ours, use our posts as a guide for doing so.