document.write( "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 \n" ); document.write( "
Algebra.Com's Answer #732686 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "The coefficients are....
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document.write( "C(n,8) = (n(n-1)...(n-6)(n-7))/8!
\r\n" ); document.write( "C(n,9) = (n(n-1)...(n-6)(n-7)(n-8))/9!
\r\n" ); document.write( "C(n,10) = (n(n-1)...(n-6)(n-7)(n-8)(n-9))/10!

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

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

\n" ); document.write( "Multiply by 2*10! and cancel the common factors n through n-7:

\n" ); document.write( "\"2%2810%29%28n-8%29+=+%289%2A10%29%2B%28n-8%29%28n-9%29\"
\n" ); document.write( "\"20n-160+=+90%2Bn%5E2-17n%2B72\"
\n" ); document.write( "\"n%5E2-37n%2B322+=+0\"
\n" ); document.write( "\"%28n-14%29%28n-23%29+=+0\"

\n" ); document.write( "The two solutions are n=14 and n=23.

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

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

\n" ); document.write( "For n=23, the coefficients are 490314, 817190, and 1144066; 817190 = (490314+1144066)/2

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