Question 1117570
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The coefficients are....<br>
C(n,4) = {{{(n(n-1)(n-2)(n-3))/24}}}<br>
C(n,5) = {{{(n(n-1)(n-2)(n-3)(n-4))/120}}}<br>
C(n,6) = {{{(n(n-1)(n-2)(n-3)(n-4)(n-5))/720}}}<br>
We need to find the value(s) of n for which C(n,5) is the arithmetic mean of C(n,4) and C(n,6).<br>
An interesting problem; but the algebra works out relatively easily....<br>
{{{(n(n-1)(n-2)(n-3)(n-4))/120 = ((n(n-1)(n-2)(n-3))/24+(n(n-1)(n-2)(n-3)(n-4)(n-5))/720)/2 = (n(n-1)(n-2)(n-3))/48+(n(n-1)(n-2)(n-3)(n-4)(n-5))/1440)}}}<br>
Multiply by the common denominator 1440 and cancel the common factors n through n-3:<br>
{{{12(n-4) = 30+(n-4)(n-5)}}}
{{{12n-48 = 30+n^2-9n+20}}}
{{{n^2-21n+98 = 0}}}
{{{(n-7)(n-14) = 0}}}<br>
The two solutions are n=7 and n=14.<br>
Check:<br>
For n=7, the coefficients are 7, 21, and 35; 21 = (7+35)/2.<br>
For n=14, the coefficients are 1001, 2002, and 3003; 2002 = (1001+3003)/2.<br>
DONE!