Coefficients in the expansion of (1+x)n in A.P, find n

[AN630078]

Homework Statement

Hello, I have question which states the coefficients of x^8,x^9 and x^10 in the binomial expansion of (1+x)^n are in an arithmetic progression. Find the possible values of n.
I think that I have been able to find a couple of similar examples but I am unable to follow their working to arrive at a solution. I become rather confused after n! is cancelled in the numerator. I have provided links to these problems. I would greatly appreciate if anyone could expand upon their working

Relevant Equations

(1+x)^n

https://www.toppr.com/ask/question/if-in-the-expansion-of-1-xn-the-coefficient-of-14th-15th-and-16th/
https://www.sarthaks.com/402983/if-...14-th-15-th-and-16-th-terms-are-in-a-p-find-n

Specifically regarding my problem the coefficients of x^8, x^9 and x^10 would be nC8,nC9,nC10.
Since they are in A.P this implies; 2 nC9=nC8+nC10
Thus, 2 n!/(n-9!)9! =n!/(n-8!)8!+n!/(n-10!)10!
Cancelling n!
2/(n-9!)9! =1/(n-8!)8!+1/(n-10!)10!

I think I need to multiply through by term(s) from the denominator but I do not know whether this is correct so far, or really how to advance any further.

 

[BvU]

Hi,

First we have to fix two errors: it is not

2 n!/(n-9!) 9! = n!/(n-8!) 8! + n!/(n-10!) 10!​

as you write, but

2 n!/((n-9)! 9!) = n!/((n-8)! 8!) + n!/((n-10)! 10!)​

rather confused after n! is canceled in the numerator.

So I suppose cancelling n! is still clear, right ? As you type, your exercise is at
$${2\over (n-9)!\,9!} ={1\over (n-8)!\,8!}+{1\over (n-10)!\,10!}\ .$$One common factor in the denominator is 8! so this simplifies to $${2\over (n-9!)\,9} ={1\over (n-8!)}+{1\over (n-10!)\,10\; 9}\ .$$Another common factor is (n-10)! so we can write $$
{2\over (n-9)\, 9} ={1\over (n-8)(n-9)}+{1\over 10\;9}$$and now you are well under way to he final quadratic equation.

 

[AN630078]

Hi,

First we have to fix two errors: it is not

2 n!/(n-9!) 9! = n!/(n-8!) 8! + n!/(n-10!) 10!​

as you write, but

2 n!/((n-9)! 9!) = n!/((n-8)! 8!) + n!/((n-10)! 10!)​

So I suppose cancelling n! is still clear, right ? As you type, your exercise is at
$${2\over (n-9)!\,9!} ={1\over (n-8)!\,8!}+{1\over (n-10)!\,10!}\ .$$One common factor in the denominator is 8! so this simplifies to $${2\over (n-9!)\,9} ={1\over (n-8!)}+{1\over (n-10!)\,10\; 9}\ .$$Another common factor is (n-10)! so we can write $$
{2\over (n-9)\, 9} ={1\over (n-8)(n-9)}+{1\over 10\;9}$$and now you are well under way to he final quadratic equation.

Thank you very much for your reply. Oh I am sorry thank you for amending my mistake.
Right, I greatly appreciate your help however I am still a bit confused. How did you identify 8! and (n-10)! as common factors?

Would the next step be to find the LCM of the denominators which would be 90(n-8)(n-9). Therefore, multiplying both sides by the LCM:
(I have attached my workings as I do not know how to write in LaTeX here and my workings look far too incomprehensible otherwise). I think I have arrived at a solution but I am not sure if this would be correct?

Thank you so much for your help

 

[BvU]

I think I have arrived at a solution but I am not sure if this would be correct?

Well, you could generate a Pascal triangle in the language of your choice ... and look at rows 14 and 23 ...

$\$

 

[BvU]

How did you identify 8! and (n-10)! as common factors?

9! = 8! * 9​

10! = 8! * 9 * 10​

​

(n-9)! = (n-10)! * (n-9)​

(n-8)! = (n-10)! * (n-9) * (n-8)​

​

$\$​

 

[AN630078]

9! = 8! * 9​

10! = 8! * 9 * 10​

​

(n-9)! = (n-10)! * (n-9)​

(n-8)! = (n-10)! * (n-9) * (n-8)​

​

$\$​

Thank you for your reply. Oh I see how silly of me, thank you for explaining how to find the common factors. Do you think that my solutions would be correct?
Thank you for the suggestion to generate a Pascal triangle, how would this help here though?

 

[BvU]

how would this help

The coefficients for the powers of 10, 9, 8 in post #4 form an arithmetic sequence, so the 14 must have been correct

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