document.write( "Question 1075272: Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.\r
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Algebra.Com's Answer #689957 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "First let's see what Pk+1 would be: \r\n" );
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document.write( "[That's always the first thing to do.  Before you start an\r\n" );
document.write( "induction proof,  you should calculate Pk+1 to see where \r\n" );
document.write( "you're headed]: \r\n" );
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document.write( "To do that, replace n by k+1 in n²-n+2 to see what the Pk+1 \r\n" );
document.write( "is, for that is what we are going for, and if we have that \r\n" );
document.write( "beforehand, we'll know when we have arrived and the proof \r\n" );
document.write( "is finished. \r\n" );
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document.write( "Substituting k+1 for n in n²-n+2,\r\n" );
document.write( "we have \r\n" );
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document.write( "Pk+1:  2 is a factor of (k+1)²-(k+1)+2 = k²+2k+1-k-1+2 = k²+k+2.  \r\n" );
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document.write( "Now that we know what Pk+1 is, we know where we're going, and \r\n" );
document.write( "we'll know we have arrived if and when we get that 2 is a factor \r\n" );
document.write( "of k²+k+2.  \r\n" );
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document.write( "Now we can start the proof: \r\n" );
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document.write( "P1:  substitute n=1 into n²-n+2 and get 1²-1+2 = 2, and 2 is \r\n" );
document.write( "indeed a factor of 2.  So P1 true. \r\n" );
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document.write( "Assume Pk:  that is, 2 is a factor of k²-k+2.  \r\n" );
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document.write( "Now if we add an even number to an even number we get an even \r\n" );
document.write( "number.  So we add the even number 2k to it and we get k²-k+2+2k \r\n" );
document.write( "or k²+k+2.  So 2 being a factor of k²-k+2 implies that 2 is a \r\n" );
document.write( "factor of k²+k+2.  \r\n" );
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document.write( "This is, Pk implies Pk+1 and P1 is true, so our induction proof is \r\n" );
document.write( "now complete.\r\n" );
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document.write( "Edwin
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