document.write( "Question 633485: USE PRINCIPLE OF MATHEMATICAL INDUCTION TO PROVE THE FORMULA
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Algebra.Com's Answer #399005 by Edwin McCravy(20056)\"\" \"About 
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1(1!)+2(2!)+3(3!)+...+n(n!) = (n+1)!-1
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document.write( "First we prove it's true for n=1\r\n" );
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document.write( "1(1!) = 1(1) = 1 and (1+1)!-1 = 2!-1 = 2-1 = 1\r\n" );
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document.write( "Now we assume it's true for n=k\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 need to show that\r\n" );
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document.write( "(2)     1(1!)+2(2!)+3(3!)+...+(k+1)(k+1)! ≟ (k+2)!-1 \r\n" );
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document.write( "We add (k+1)(k+1)! to both sides of (1)\r\n" );
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document.write( "(1)     1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)! = (k+1)!-1+(k+1)(k+1)! =\r\n" );
document.write( "                                                = (k+1)!+(k+1)(k+1)!-1 =\r\n" );
document.write( "                                                = (k+1)![1+(k+1)]-1 =\r\n" );
document.write( "                                                = (k+1)![1+k+1]-1 =\r\n" );
document.write( "                                                = (k+1)!(k+2)-1 =\r\n" );
document.write( "                                                = (k+2)!-1\r\n" );
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document.write( "So the truth of (1) implies the truth of (2). So the induction is complete.\r\n" );
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document.write( "Edwin
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