document.write( "Question 924749: 5) Find the value of S=(1/1!)+(3/2!)+(7/3!)+(13/4!)+(21/5!)+(31/6!)+(43/7!)+......
\n" ); document.write( "[Given that e^x = 1+(x/1!)+(x^2/2!)+.....]
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Algebra.Com's Answer #561578 by Edwin McCravy(20056) $\"\"$ $\"About$

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S=(1/1!)+(3/2!)+(7/3!)+(13/4!)+(21/5!)+(31/6!)+(43/7!)+......
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document.write( "given:  (I hope you understand sigma summation notation)\r\n" );
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document.write( "Therefore\r\n" );
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document.write( "We need first to find the nth term of the sequence of the numerators,\r\n" );
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document.write( "The sequence of numerators is 1,3,7,13,21,31,43,... \r\n" );
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document.write( "They are all odd. Perhaps if we subtract 1 from each, \r\n" );
document.write( "we might recognize a pattern,\r\n" );
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document.write( "That would be the sequence 0,2,6,12,20,30,42.  Aha! That\r\n" );
document.write( "pattern is 1*0,2*1,3*2,4*3,5*4,6*5,7*6 so its nth term is n(n-1)\r\n" );
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document.write( "So the nth term of the numerators is 1 more than that, or n(n-1)+1 \r\n" );
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document.write( "and\r\n" );
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document.write( " becomes\r\n" );
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document.write( "That has an n(n-1) term on top, and we notice that if n≧2, n! can be written\r\n" );
document.write( "n(n-1)(n-2)! which has that factor.  So let's write out the first term of S,\r\n" );
document.write( "which is  so we can start the sum at n=2 instead of n=1, so,\r\n" );
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document.write( "(1)   \r\n" );
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document.write( "The first summation in (1) above:\r\n" );
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document.write( "we substitute n-2=k and n=k+2, and it becomes\r\n" );
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document.write( "The other summation in (1) above:\r\n" );
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document.write( " is the summation for e except for the first two terms \r\n" );
document.write( "where n=0 and n=1.  Therefore,\r\n" );
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document.write( "Therefore (1) above becomes\r\n" );
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document.write( "(1)   1 + e + e-2 = 2e-1\r\n" );
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document.write( "The correct choice is C.\r\n" );
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document.write( "Edwin

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