Question 1102843
Proof by induction (using C(n,r) notation):

n=0:   C(0,0) = 1 = C(1,1)

Assume its true for n=k:
n=k:   C(r,r)+C(r+1,r)+ … + C(k-1,r) + C(k,r) = C(k+1, r+1)    (*) 
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Now we need to show that it follows that (*) leads to the equality holding for n=k+1:
n=k+1:   C(r,r)+C(r+1,r)+ … + C(k-1,r) + C(k,r) + C(k+1,r) 
          
Adding brackets around all but the last term from above:   
[ <b>C(r,r)+C(r+1,r)+ … + C(k-1,r) + C(k,r) </b>] + C(k+1,r) 

The part between brackets [ ] is  (*):
= C(k+1, r+1) + C(k+1, r)
= {{{  ( (k+1)!/((k-r)!(r+1)!) ) + (k+1)!/((k+1-r)!(r!)) }}} 
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— re-writing the above by multiplying the 2nd term by (r+1)/(r+1) and 
— pulling out the first factor of  (k+1-r)! in the denominator:
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= {{{  ( (k+1)!/((k-r)!(r+1)!) ) + (k+1)!*(r+1)/((k+1-r)(k-r)!(r+1)!))) }}} 
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— allows us to factor to:
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= {{{  ( (k+1)!/((k-r)!(r+1)!) ) * ( 1 + (r+1)/(k+1-r) ) }}}
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= {{{  ( (k+1)!/((k-r)!(r+1)!) ) * ( ( k+1-r + (r+1)) /(k+1-r) ) }}}
= {{{  ( (k+1)!/((k-r)!(r+1)!) ) * ( (k+2)/(k-r+1) ) }}}
= {{{   (k+2)!/((k+1-r)!(r+1)!) }}}
= {{{  C(k+2,r+1) }}}   Done.  Assuming truth for n=k ==> true for n=k+1
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The last statement can be re-written as
= {{{  C(n+1,r+1) }}}    
because n=k+1.