document.write( "Question 1025556: Use mathematical induction to prove the statement is true for all positive integers n.\r
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Algebra.Com's Answer #640831 by Edwin McCravy(20056)\"\" \"About 
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document.write( "n³+2n\r\n" );
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document.write( "If n=1 then 1³+2(1) = 1+2 = 3 which is divisible by 3.\r\n" );
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document.write( "Let us assume that n=k is some integer (perhaps 1) such that\r\n" );
document.write( "k³+2k is divisible by n.  That is, there is some positive \r\n" );
document.write( "integer p such that k³+2k = 3p \r\n" );
document.write( "[In the case where n=1, then p=1]\r\n" );
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document.write( "We want to show that the expression n³+2n with k+1 substituted\r\n" );
document.write( "for n also gives a multiple of 3.\r\n" );
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document.write( "We examine the case where n=k+1 and multiply it all the way out:\r\n" );
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document.write( "(k+1)³+2(k+1) = k³+6k²+11k+6.\r\n" );
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document.write( "We notice that this differs from k³+2k by 6k²+9k+6.\r\n" );
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document.write( "So we add 6k²+9k+6 to both side of\r\n" );
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document.write( "k³+2k = 3p\r\n" );
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document.write( "and get\r\n" );
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document.write( "k³+2k+6k²+9k+6 = 3p+6k²+9k+6 = 3(p+2k²+3k+2)\r\n" );
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document.write( "So (k+1)³+2(k+1) = 3(p+2k²+3k+2)\r\n" );
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document.write( "which is a multiple of 3, so the theorem is proved.\r\n" );
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document.write( "Edwin
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