Binomial expansion of term with x^2

[Roodles01]

I have to determine the coefficient of an x term in an expansion such as this;
Determine the coefficient of x^18 in the expansion of (1/14 x^2 -7)^16

The general term in the binomial expansion is
nCk a^k b^(n−k)
I could let
a = (1/14 x^2)
b = -7
n = 16
k = 9?

I have no real idea of how to go about finding this coefficient using the binomial theorem.

Having expanded the expression to the 10th term I get

8C9 (-7) (1/14 x^2)^9

I'm using nCk = n! / (n-k)!k! but can't evaluate this as it is a negative.

I'm assuming that the 8C9 bit is just the opposite of 6th term i.e. 12C5 = 792 (looking at Pascal's triangle this is on the opposite side), but I can get the x^18 bit (I'm assuming the (x^2)^9 can be x^18 here)

Can someone check, please?

 

[Ray Vickson]

I have to determine the coefficient of an x term in an expansion such as this;
Determine the coefficient of x^18 in the expansion of (1/14 x^2 -7)^16

The general term in the binomial expansion is
nCk a^k b^(n−k)
I could let
a = (1/14 x^2)
b = -7
n = 16
k = 9?

I have no real idea of how to go about finding this coefficient using the binomial theorem.

Having expanded the expression to the 10th term I get

8C9 (-7) (1/14 x^2)^9

I'm using nCk = n! / (n-k)!k! but can't evaluate this as it is a negative.

I'm assuming that the 8C9 bit is just the opposite of 6th term i.e. 12C5 = 792 (looking at Pascal's triangle this is on the opposite side), but I can get the x^18 bit (I'm assuming the (x^2)^9 can be x^18 here)

Can someone check, please?

Conventionally, nCk for integers 0 < n < k is regarded as zero. Why did you write 8C9?

RGV

 

[HallsofIvy]

If you let y= x^2, then you are looking for the coefficient of y^9 in ((1/14)y- 7)^16.
That will be, of course, 16C9 (1/14)^9(-7)^7

 

[Roodles01]

Of course you are right HallsofIvy.

I was doing this

(OK I put the 6th term when I meant 7th)
I put 8C9 as I understand that when I expand the expression (1/14 x^2 - 7) longhand I get;

-7^16 + (16C1*-7^15*(1/14 x^2)) + (15C2*-7^14*(1/14 x^2)) + . . . . + (11C6*-7^10*(1/14 x^2)) + . . . . . . . . + (8C9*-7^7*(1/14 x^2)) . . . .

Where (11C6*-7^10*(1/14 x^2)) is the 7th term of the expansion
& (8C9*-7^7*(1/14 x^2)) is the 10th term

Pascals triangle, being symmetrical, should reflect the coefficients around the8th & 9th terms. So I'm assuming that the coefficient of the 7th term should be the same as the 10th term.

When I try to calculate 8C9 (which I did longhand, shown above)
I use form nCk = n! / (n-k)!*k!
If I let n=8
& k=9
I get 8!/(-1)!*9!

I work this out to be;
40320 / -1 * 362880 = -0.1

I know this isn't right so why did I do this wrong!

 

[HallsofIvy]

[itex]_8C_9[/itex] would be the coefficient of [itex]a^9[/itex] in [itex](a+ b)^8[/itex] and there is no such term! Do you mean [itex]_9C_8[/itex]?