document.write( "Question 1163040: If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1. \n" ); document.write( "
Algebra.Com's Answer #786992 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Something is wrong with this problem, as I show below.\r\n" );
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\n" ); document.write( "If the series {ak} satisfies that a1 = 1, and a2 = 2, and ak - 4a(k-1) + 3a(k-2) = 0 (k >= 0), then ak = (1 + p)/q. Find p and q for all values of k >= 1.
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document.write( "We substitute k=2\r\n" );
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document.write( "\"a%5B2%5D+-+4a%5B1%5D+%2B+3a%5B0%5D+=+0\"\r\n" );
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document.write( "\"2+-+4%281%29+%2B+3a%5B0%5D+=+0\"\r\n" );
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document.write( "\"2+-+4+%2B+3a%5B0%5D+=+0\"\r\n" );
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document.write( "-----\r\n" );
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document.write( "Next we substitute k=3\r\n" );
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document.write( "\"a%5B3%5D+-+4a%5B2%5D+%2B+3a%5B1%5D+=+0\"\r\n" );
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document.write( "\"a%5B3%5D+-+4%282%29+%2B+3%281%29+=+0\"\r\n" );
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document.write( "\"a%5B3%5D+-+8+%2B+3+=+0\"\r\n" );
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document.write( "So we have these two equations in p and q:\r\n" );
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document.write( "\"system%282q=1%2Bp%2C5q=1%2Bp%29\"\r\n" );
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document.write( "This can't be because the solution to that system is p=-1, q=0,\r\n" );
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document.write( "But q is in the denominator of \r\n" );
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document.write( "\"a%5Bk%5D+=+%281+%2B+p%29%2Fq\"\r\n" );
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document.write( "and so q cannot be 0.  So the problem is botched.\r\n" );
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document.write( "You can correct it in the space below if you like\r\n" );
document.write( "and I'll get back to you by email. \r\n" );
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document.write( "Edwin
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