Question 924749
S=(1/1!)+(3/2!)+(7/3!)+(13/4!)+(21/5!)+(31/6!)+(43/7!)+......
<pre>
given:  (I hope you understand sigma summation notation)

{{{e^x}}}{{{""=""}}}{{{matrix(1,2,sum("",n=0,infinity),(x^n/n!))}}}

Therefore

{{{e=e^1}}}{{{""=""}}}{{{matrix(1,2,sum("",n=0,infinity),(1/n!))}}}

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We need first to find the nth term of the sequence of the numerators,

The sequence of numerators is 1,3,7,13,21,31,43,... 

They are all odd. Perhaps if we subtract 1 from each, 
we might recognize a pattern,

That would be the sequence 0,2,6,12,20,30,42.  Aha! That
pattern is 1*0,2*1,3*2,4*3,5*4,6*5,7*6 so its nth term is n(n-1)

So the nth term of the numerators is 1 more than that, or n(n-1)+1 

and

{{{S=(1/1!)+(3/2!)+(7/3!)+(13/4!)+(21/5!)+(31/6!)+(43/7!)+""*""*""*""}}} becomes

{{{matrix(1,3,S,""="",matrix(1,2,sum("",n=1,infinity),( (n(n-1)+1)/n!) ))  }}}

That has an n(n-1) term on top, and we notice that if n&#8807;2, n! can be written
n(n-1)(n-2)! which has that factor.  So let's write out the first term of S,
which is {{{(1(1-1)+1)/1!=1}}} so we can start the sum at n=2 instead of n=1, so,

{{{matrix(1,3,S,""="",matrix(1,2,sum("",n=1,infinity), ((n(n-1)+1)/n! )))  }}}{{{""=""}}}{{{1+matrix(1,2,sum("",n=2,infinity), (n(n-1)+1)/n! )  }}}{{{""=""}}}{{{1+matrix(1,2,sum("",n=2,infinity), ( (n(n-1)+1)/( n*(n-1)*(n-2)!  )  ))  }}}
{{{""=""}}}{{{1}}}{{{""+""}}}{{{matrix(1,2,sum("",n=2,infinity), ( (n(n-1))/( n*(n-1)*(n-2)!  )  ))  }}}{{{""+""}}}{{{matrix(1,2,sum("",n=2,infinity), ( 1/( n*(n-1)*(n-2)!)))}}}{{{""=""}}}
{{{1}}}{{{""+""}}}{{{matrix(1,2,sum("",n=2,infinity), ( (cross(n)(cross(n-1)))/( cross(n)*(cross(n-1))*(n-2)!  )  ))  }}}{{{""+""}}}{{{matrix(1,2,sum("",n=2,infinity), ( 1/(n!)))}}}{{{""=""}}}

(1)   {{{1}}}{{{""+""}}}{{{matrix(1,2,sum("",n=2,infinity), ( 1/(n-2)!    ))  }}}{{{""+""}}}{{{matrix(1,2,sum("",n=2,infinity), ( 1/( n!)))}}}

The first summation in (1) above:

{{{matrix(1,2,sum("",n=2,infinity),(1/(n-2)!))}}}

we substitute n-2=k and n=k+2, and it becomes

{{{matrix(1,2,sum("",k+2=2,infinity),(1/(k)!))}}}{{{""=""}}}{{{matrix(1,2,sum("",k=0,infinity),(1/k!))}}}{{{""=""}}}{{{e}}}.

The other summation in (1) above:

{{{matrix(1,2,sum("",n=2,infinity), ( 1/( n!)))}}} is the summation for e except for the first two terms 
where n=0 and n=1.  Therefore,

{{{matrix(1,2,sum("",n=2,infinity),(1/(n!)))}}}{{{""=""}}}{{{matrix(1,2,sum("",n=0,infinity),(1/(n!)))-1/0!-1/1!}}}{{{""=""}}}{{{e-1-1}}}{{{""=""}}}{{{e-2}}}

Therefore (1) above becomes

(1)   1 + e + e-2 = 2e-1

The correct choice is C.

Edwin</pre>