document.write( "Question 172778: The first three terms in the expansion of (1+ay)^n are 1,12y, and 68y^2\r
\n" ); document.write( "\n" ); document.write( "Evaluate a and n. Use the fact that:\r
\n" ); document.write( "\n" ); document.write( "(1+ay)^n=1+nay+n(n-1)/2(ay)^2+...\r
\n" ); document.write( "\n" ); document.write( "my teacher told me to use: c(n,k)(1)^n(ay)^k to figure it out and to refer to pascals triangle, But it hasnt helped. I would very much appricate if someone could help.
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Algebra.Com's Answer #127654 by vleith(2983)\"\" \"About 
You can put this solution on YOUR website!
First three terms are 1, 12y and 68y^2
\n" ); document.write( "Your hint says (1+ay)^n=1+nay+n(n-1)/2(ay)^2+
\n" ); document.write( "so (1+ay)^n=1+12y+68y^2+...
\n" ); document.write( "12y=nay, so na=12 and a=12/n
\n" ); document.write( "68y^2 = (n(n-1)/2)(ay)^2 = (n(n-1)/2)a^2y^2. so (n(n-1)/2)a^2 = 68\r
\n" ); document.write( "\n" ); document.write( "Now substitute the value for a into the second equation and solve
\n" ); document.write( "\"n%28n-1%29a%5E2%2F2+=+68\"
\n" ); document.write( "\"%28%28n%5E2-n%29%2F2%29+%2A+%2812%2Fn%29%5E2+=+68\"
\n" ); document.write( "\"%28n%5E2-n%29%28144%2F2%29+%2F+%28n%5E2%29+=+68+\"
\n" ); document.write( "\"%28n%5E2+-n%2972+=+68n%5E2\"
\n" ); document.write( "\"72n%5E2+-+72n+=+68n%5E2\"
\n" ); document.write( "\"4n%5E2+=+72n\"
\n" ); document.write( "\"4n+=+72\"
\n" ); document.write( "\"n+=+18\"
\n" ); document.write( "so a = 2/3\r
\n" ); document.write( "\n" ); document.write( "now check against the hint to see if the values work
\n" ); document.write( "a*n = (2/3)*18 = 12. so far so good
\n" ); document.write( "(18*17/2)*(4/9) = 68\r
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