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If n is an odd natural number, then sum...

If n is an odd natural number, then `sum_(r=0)^n (-1)^r/(nC_r)` is equal to

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We have , `sum_(r=0)^(n)((-1)^(r))/(""^(n)C_(r))= sum_(r=0)^(n) [((-1)^(r))/(""^(n)C_(r)) + ((-1)^(n-r))/(""^(n)C_(n-r)) ]`
`sum_(r=0)^((n+1)/(2))(-1)^(r) = [(-1)/(""^(n)C_(r)) + ((-1)^(n))/(""^(n)C_(n-r)) ]=sum_(r=0)^((n+1)/(2))(-1)^(r) = [(1)/(""^(n)C_(r)) + (1)/(""^(n)C_(r)) ]= 0`
` [ because " n is odd and " ^(n)CC_(r) = ""^(n)C_(n-r)]` .
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