S.O.S. Mathematics CyberBoard

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 it is a positive integer.

Is it possible to do it manually instead of using a tables book?

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Yes it is - you have to set nC9 = (nC8 + nC10)/2 and solve for n.
This equation, when you apply the definition of nCr (binomial theorem), leads to a quadratic in n with two integer solutions. You will have to be adept at simplifying with factorials, but the solutions are relatively small integers. Good luck!

Sorry but you missed out another 8 possibile values of n.

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Are you refering to the trivial coefficients of zero for n less than 8? That's "clever" - I did miss them, but there are only 7, as n is to be positive. What would be the 8th "other solution" and what method is needed to find it?

Of course one could argue that a coefficient of zero really means the poynomial doesn't actually HAVE that term!

is nc9 = n(n-1)(n-2)(n-3)(n-4)....(n-8)? If so, how do I simplify them (including nC8 and nC10) ?

nCr = n!/[r!(n-r)!]