document.write( "Question 1040101: In the expansion of (1+x)n, two times the coefficient of x5 is equal to the sum of the coefficient of x4 + x6. Find the possible values of n.\r
\n" ); document.write( "\n" ); document.write( "2(nC5) = nC4 + nC6 , find n \n" ); document.write( "
Algebra.Com's Answer #654916 by Boreal(15235)\"\" \"About 
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2*(n)(n-1)(n-2)(n-3)(n-4)/5! (the n-5)! cancels the (n-5)! on the bottom
\n" ); document.write( "That equals n(n-1)(n-2)(n-3)/4! +n(n-1)(n-2)(n-3(n-4)(n-5)/6!
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\n" ); document.write( "I can cancel n(n-1)(n-2)(n-3) from everything, because it is in all terms.
\n" ); document.write( "2(n-4)/120=1/24 + (n-4)(n-5)/720
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\n" ); document.write( "The first term will be (n-4)/60
\n" ); document.write( "Now put everything over a common denominator of 720
\n" ); document.write( "12(n-4)=30+(n-4)(n-5), and since they are all over the same number 720, I can remove it without changing the result.
\n" ); document.write( "12n-48=30+n^2-9n+20
\n" ); document.write( "0=n^2-21n+98
\n" ); document.write( "0=(n-14)(n-7)
\n" ); document.write( "n=14 or 7
\n" ); document.write( "14C5=2002
\n" ); document.write( "14C4=1001
\n" ); document.write( "14C6=3003
\n" ); document.write( "7C5=21
\n" ); document.write( "7C4=35
\n" ); document.write( "7C6=7
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