document.write( "Question 804680: Prove that for all positive integers n,
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Algebra.Com's Answer #484902 by Edwin McCravy(20055)\"\" \"About 
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document.write( "Prove:\r\n" );
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document.write( "1•1!+2•2!+3•3!+...+n•n! = (n+1)!-1\r\n" );
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document.write( "The proof is by induction.\r\n" );
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document.write( "It is true for n=1\r\n" );
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document.write( "1•1! = 1 and (1+1)!-1 = 2!-1 = 2-1 = 1\r\n" );
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document.write( "Assume k is such that the proposition is true for n ≦ k\r\n" );
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document.write( "We are assuming that \r\n" );
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document.write( "(1)   1•1!+2•2!+3•3!+...+k•k! = (k+1)!-1\r\n" );
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document.write( "We must use (1) to show (2) \r\n" );
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document.write( "(2)   1•1!+2•2!+3•3!+...+(k+1)•(k+1)! = ((k+1)+1)!-1 = (k+2)!-1\r\n" );
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document.write( "We start with\r\n" );
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document.write( "(1)   1•1!+2•2!+3•3!+...+k•k! = (k+1)!-1\r\n" );
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document.write( "We add (k+1)•(k+1)! to both sides\r\n" );
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document.write( "      1•1!+2•2!+3•3!+...+k•k!+(k+1)•(k+1)! = (k+1)!-1+(k+1)•(k+1)!\r\n" );
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document.write( "= (k+1)!+(k+1)•(k+1)!-1 =\r\n" );
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document.write( "Factor (k+1)! out of the first two terms:\r\n" );
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document.write( "(k+1)!•[1+(k+1)]-1 = (k+1)!•[1+k+1]-1 = (k+1)!•(k+2)-1 = (k+2)!-1\r\n" );
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document.write( "So we have shown (2)\r\n" );
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document.write( "We know that it is true for n=1, and for n=k=1 we know that it is true for\r\n" );
document.write( "n=k=2.  That means we know it is true for n=k=3, and that means we know it\r\n" );
document.write( "is true for n=k=4, and so on.\r\n" );
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document.write( "QED\r\n" );
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document.write( "Edwin
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