Solving for nC8,nC9, and nC10 in an Arithmetic Progression

[AN630078]

Homework Statement

Hello, I have a question concerning finding the coefficients of x^8,x^9 and x^10 in the binomial expansion of (1+x)^n if they are in an arithmetic progression. Find the possible values of n.

Relevant Equations

(1+x)^n=1+nx+n(n-1)/2!x^2+...

Well, I am having a little difficulty knowing how to approach finding a solution to this problem. I am aware that in an arithmetic progression the first term is a and there is a constant common difference defined as d=un+1-un

Expanding the binomial given;

(1+x)^n=1+nx+n(n-1)/2!x^2+n(n-1)(n-2)/3!x^3+...n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)/8!x^8+n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)/9!x^9+n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)/10!x^10

Therefore, the coefficients ofx^8,x^9 and x^10 are n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)/8!,n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)/9! andn(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)/10! respectively.

Does this imply that nC8,nC9 and nC10 are in an A.P. So that 2nC9=nC8+nC10

Thus, 2=nC8/nC9+nC10/nC9

I do not know whether this is a suitable start to finding a solution or necessarily how to progress any further. I would be very grateful for any suggestions.

 

[anuttarasammyak]

$$(1+x)^n=(1+x)(1+x)(1+x)...(1+x)$$
In order to know coefficient of $x^m$ of m<n in expansion, consider the number of ways you choose m brackets from which you pick up x not 1 for multiplication among all the n brackets.

 

[Stephen Tashi]

Does this imply that nC8,nC9 and nC10 are in an A.P. So that 2nC9=nC8+nC10

Yes

Thus, 2=nC8/nC9+nC10/nC9

I do not know whether this is a suitable start to finding a solution or necessarily how to progress any further. I would be very grateful for any suggestions.

Write expressions of the form "nCr" as n!/((n−r)!r!). This allows the fractions involved to be simplified.

For example ( n!/ ((n−r)!r!) / (n!/((n−(r−1)!(r−1)!) =((n−(r−1))!(r−1)!)/ ((n−r)!r!)

 

[Stephen Tashi]

I'm having trouble getting the forum's LaTex to work.

I'll try again:

For example:

$\frac{\frac{n!}{(n-(r+1))!(r+1)!} }{\frac{n!}{(n-r)! r!} }$

$= \frac{(n-r)!}{(n-(r+1))!} \frac{r!}{(r+1)!)}$

$= (n-r) \frac{1}{(r+1)}$