SOLUTION: if the coefficient of x^8, x^9 and x^10 in the expansion of (1+x)^n are in arithmetic progression, find the values of n where n is a positive integer

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Question 1117569: if the coefficient of x^8, x^9 and x^10 in the expansion of (1+x)^n are in arithmetic progression, find the values of n where n is a positive integer
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The coefficients are....
C(n,8) = (n(n-1)...(n-6)(n-7))/8!

C(n,9) = (n(n-1)...(n-6)(n-7)(n-8))/9!

C(n,10) = (n(n-1)...(n-6)(n-7)(n-8)(n-9))/10!

We need to find the value(s) of n for which the three coefficients are in arithmetic progression -- i.e., for which C(n,9) is the arithmetic mean of C(n,8) and C(n,10).

An interesting problem; but the algebra works out relatively easily....
(n(n-1)...(n-6)(n-7)(n-8))/9! = ((n(n-1)...(n-6)(n-7))/8!+(n(n-1)...(n-6)(n-7)(n-8)(n-9))/10!)/2

Multiply by 2*10! and cancel the common factors n through n-7:

2%2810%29%28n-8%29+=+%289%2A10%29%2B%28n-8%29%28n-9%29
20n-160+=+90%2Bn%5E2-17n%2B72
n%5E2-37n%2B322+=+0
%28n-14%29%28n-23%29+=+0

The two solutions are n=14 and n=23.

Check:

For n=14, the coefficients are 1001, 2002, and 3003; 2002 = (1001+3003)/2.

For n=23, the coefficients are 490314, 817190, and 1144066; 817190 = (490314+1144066)/2

DONE!