document.write( "Question 976324: If the series {ak} satisfies a1=1, a2=2, and (ak)-4(ak-1)+3(ak-2)=0 for K greater than or equal to 3, then ak=(1+p)/q for k greater than or equal to 1. Find p and q.
\n" ); document.write( "Please use the below representation-
\n" ); document.write( "ak = kth term = a subscript k
\n" ); document.write( "a1 = 1st term = a subscript 1
\n" ); document.write( "a2 = 2nd term = a subscript 2
\n" ); document.write( "ak-1 = (k-1)th term = a subscript k-1
\n" ); document.write( "ak-2 = (k-2)th term = a subscript k-2\r
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Algebra.Com's Answer #597930 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
If the series {ak} satisfies a1=1, a2=2, and (ak)-4(ak-1)+3(ak-2)=0 for K greater than or equal to 3, then ak=(1+p)/q for k greater than or equal to 1. Find p and q.
\n" ); document.write( "Please use the below representation-
\n" ); document.write( "ak = kth term = a subscript k
\n" ); document.write( "a1 = 1st term = a subscript 1
\n" ); document.write( "a2 = 2nd term = a subscript 2
\n" ); document.write( "ak-1 = (k-1)th term = a subscript k-1
\n" ); document.write( "ak-2 = (k-2)th term = a subscript k-2
\n" ); document.write( "------
\n" ); document.write( "a(3) - 4*a(2) + 3*a(1) = 0
\n" ); document.write( "---
\n" ); document.write( "a(3) - 4*2 + 3*1 = 0
\n" ); document.write( "---
\n" ); document.write( "a(3) - 8 + 3 = 0
\n" ); document.write( "a(3) = 5
\n" ); document.write( "------------
\n" ); document.write( "ak=(1+p)/q
\n" ); document.write( "Therefore::
\n" ); document.write( "a(3) = (1+p)/q = 5/1
\n" ); document.write( "------
\n" ); document.write( "1+p = 5, so p = 4
\n" ); document.write( "q = 1
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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