document.write( "Question 1117570: In the expansion of (1+x)^n, the coefficient of x^5 is the arithmetic mean of the coefficients of x^4 and x^6. calculate the possible values of n \n" ); document.write( "
Algebra.Com's Answer #732684 by greenestamps(13215)\"\" \"About 
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\n" ); document.write( "The coefficients are....

\n" ); document.write( "C(n,4) = \"%28n%28n-1%29%28n-2%29%28n-3%29%29%2F24\"

\n" ); document.write( "C(n,5) = \"%28n%28n-1%29%28n-2%29%28n-3%29%28n-4%29%29%2F120\"

\n" ); document.write( "C(n,6) = \"%28n%28n-1%29%28n-2%29%28n-3%29%28n-4%29%28n-5%29%29%2F720\"

\n" ); document.write( "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).

\n" ); document.write( "An interesting problem; but the algebra works out relatively easily....

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\n" ); document.write( "Multiply by the common denominator 1440 and cancel the common factors n through n-3:

\n" ); document.write( "\"12%28n-4%29+=+30%2B%28n-4%29%28n-5%29\"
\n" ); document.write( "\"12n-48+=+30%2Bn%5E2-9n%2B20\"
\n" ); document.write( "\"n%5E2-21n%2B98+=+0\"
\n" ); document.write( "\"%28n-7%29%28n-14%29+=+0\"

\n" ); document.write( "The two solutions are n=7 and n=14.

\n" ); document.write( "Check:

\n" ); document.write( "For n=7, the coefficients are 7, 21, and 35; 21 = (7+35)/2.

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

\n" ); document.write( "DONE!
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