If ` a_(n) = sum_(r=0)^(n) (1)/(""^(n)C_(r)) ` , find the
value of ` sum_(r=0)^(n) (r)/(""^(n)C_(r))`

Explanation

Let `P sum_(r=0)^(n) (r)/(""^(n)C_(r))` …(i)
Replacing by (n-r) in Eq . (i) , we get
`P= sum_(r=0)^(n) (n-r)/(""^(n)C_(n-r)) = sum_(r=0)^(n) ((n-r))/(""^(n)C_(r))" " [because ""^(n)C_(r) =""^(n)C_(n-r) ]...(ii)`
On adding Eqs. (i) and (ii) , we get
`P= sum_(r=0)^(n) (n-r)/(""^(n)C_(n-r)) = sum_(r=0)^(n) (1)/(""^(n)C_(r)=) = na_(n)" " ` [ given]
` therefore P = (n)/(2) a_(n)`
Hence , ` sum_(r=0)^(n) (r)/(""^(n)C_(r)) = (n)/(2) a_(n)`