In the expansion of (1+x)^n the coefficient of x^9 is the arithmetic mean of the coefficient of x^8 and x^10. Find the possible values of n where n>0

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The coefficient of the kth power of x in the expansion (1 + x)^n is n!/(n-k)!*k!

Here, in the expansion of (1 + x)^n, the coefficient of x^9 is the arithmetic mean of the coefficient of x^8 and x^10. This gives:

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

=> 1/(n-8)!*8! + 1/(n - 10)!*10! = 2/(n - 9)!*9!

=> 1/(n-8)! + 1/(n - 10)!*9*10 = 2/(n - 9)!*9

=> 90/(n - 8)! + (n - 8)(n - 9)/(n - 8)! = 20(n - 8)/(n - 8)!

=> 90 + (n - 8)(n - 9) = 20(n - 8)

=> 250 + (n - 8)(n - 9) = 20n

=> 250 + n^2 - 17n + 72 = 20n

=> n^2 - 37n + 322 = 0

=> n^2 - 14n - 23n + 322 = 0

=> n(n - 14) - 23(n - 14) = 0

=> (n - 14)(n - 23) = 0

n = 14 and n = 23

**The required condition is met for n = 14 and n = 23**

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

=> 1/(n-8)!*8! + 1/(n - 10)!*10! = 2/(n - 9)!*9!"

can i ask something?

Why is it that the n! from the respective numerators can be removed?

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