document.write( "Question 929683: prove that the sum of all natural numbers divisible by 5 is equal to 5n(n+1)/2 using mathematical induction. \n" ); document.write( "
Algebra.Com's Answer #564646 by Edwin McCravy(20055)\"\" \"About 
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prove that the sum of all natural numbers divisible by 5 is equal to 5n(n+1)/2 using mathematical induction.
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document.write( "First we want to show that \r\n" );
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document.write( "if n were some integer for which the formula \"5n%28n%2B1%29%2F2\" gives the sum of\r\n" );
document.write( "the first n positive multiples of 5, from 5 through 5n, then the same formula\r\n" );
document.write( "with (n+1) substituted for n, would give the sum of the first n+1 positive\r\n" );
document.write( "multiples of 5, from 5 up to 5(n+1).\r\n" );
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document.write( "Let's hypothetically suppose that n is some integer for which the formula\r\n" );
document.write( "\"5n%28n%2B1%29%2F2\" gives the sum of the first n positive multiples of 5\r\n" );
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document.write( "If that were the case then if we added 5(n+1) to it we would have the sum\r\n" );
document.write( "of the first n+1 multiples of 5.  So let's add 5(n+1) to the formula to see if\r\n" );
document.write( "we get the formula \"5n%28n%2B1%29%2F2\" with n+1 substituted for n.\r\n" );
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document.write( "When we add them we get:\r\n" );
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document.write( " \"5n%28n%2B1%29%2F2\"\"%22%22%2B%22%22\"\"5%28n%2B1%29\"\r\n" );
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document.write( "We get an LCD:\r\n" );
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document.write( " \"5n%28n%2B1%29%2F2\"\"%22%22%2B%22%22\"\"10%28n%2B1%29%2F2\"\r\n" );
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document.write( "\"%285n%28n%2B1%29%2B10%28n%2B1%29%29%2F2\"\r\n" );
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document.write( "We factor out (n+1) on top:\r\n" );
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document.write( " \"%28%28n%2B1%29%285n%2B10%29%29%2F2\"\r\n" );
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document.write( "We factor out 5 from the second paretheses on top:\r\n" );
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document.write( " \"%28%28n%2B1%29%285%29%28n%2B2%29%29%2F2\"\r\n" );
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document.write( " \"%285%28n%2B1%29%28n%2B2%29%29%2F2\"\r\n" );
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document.write( "And now we see if this is what we'd get if we substituted n+1 for \r\n" );
document.write( "n in the formula \"5n%28n%2B1%29%2F2\", so we substitute n+1 for n in it\r\n" );
document.write( "to see:\r\n" );
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document.write( "\"%285%28n%2B1%29%28%28n%2B1%29%5E%22%22%2B1%29%29%2F2\"\r\n" );
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document.write( "\"%285%28n%2B1%29%28n%2B1%2B1%29%29%2F2\"\r\n" );
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document.write( "\"%285%28n%2B1%29%28n%2B2%29%29%2F2\"\r\n" );
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document.write( "Yes that is what we got.\r\n" );
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document.write( "So therefore if we can find a value of n for which the formula\r\n" );
document.write( "\"%285%28n%2B1%29%28n%2B2%29%29%2F2\" will give the sum of the first n positive\r\n" );
document.write( "multiples of 5, then it will also give us the sum of the first\r\n" );
document.write( "n+1 positive multiples of 5.\r\n" );
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document.write( "So all that's left is to find such a value of n.\r\n" );
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document.write( "Let's try n=1 to see if the formlaq holds for that:\r\n" );
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document.write( "\"5n%28n%2B1%29%2F2\"\r\n" );
document.write( "\"5%281%29%281%2B1%29%2F2\"\r\n" );
document.write( "\"5%282%29%2F2\"\r\n" );
document.write( "\"10%2F2\"\r\n" );
document.write( "\"5\"\r\n" );
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document.write( "Indeed it works for n=5 because the first positive multiple of 5 is 5.\r\n" );
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document.write( "So therefore since the formula holds when n=1, from what we proved above,\r\n" );
document.write( "the formula will also hold when n=1+1=2.\r\n" );
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document.write( "So therefore since the formula holds when n=2, from what we proved above,\r\n" );
document.write( "the formula will also hold when n=+1=3.\r\n" );
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document.write( "So therefore since the formula holds when n=3, from what we proved above,\r\n" );
document.write( "the formula will also hold when n=3+1=4.\r\n" );
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document.write( "etc., etc., etc.,\r\n" );
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document.write( "This never stops, so the formula holds for all integers n.\r\n" );
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document.write( "[Many books teach you to do the second part first and the first part second, \r\n" );
document.write( "so you can reverse them if you like.]\r\n" );
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document.write( "Edwin

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