If n is an even natural number , then `sum_(r=0)^(n) (( -1)^(r))/(""^(n)C_(r))` equals

Explanation

We have `sum_(r=0)^(n) ((-1)^(r))/(""^(n)C_(r)) sum_(r=0)^((n)/(2)-1) = [((-1)^(r))/(""^(n)C_(r)) + ((-1)^(n-r))/(""^(n)C_(n-r)) ] + ((-1)^(n//2))/(""^(n)C_(n//2))`
` =sum_(r=0)^((n)/(2)-1)(-1)^(r) = [(1)/(""^(n)C_(r)) + ((-1)^(n))/(""^(n)C_(r)) ]+((-1)^(n//2))/(""^(n)C_(n//2))`
` =sum_(r=0)^((n)/(2)-1)(-1)^(r) = [(1)/(""^(n)C_(r)) + (1)/(""^(n)C_(r)) ]+((-1)^(n//2))/(""^(n)C_(n//2))`
` = [sum_(r=0)^((n)/(2)-1)(-1)^(r)*(2)/(""^(n)C_(r))] + ((1-)^(n//2))/(""(n)C_(n//2))`