If ` (1 + x)^(n) = sum_(r=0)^(n) C_(r)x^(r)`, then prove that
`(sumsum)_(0leiltjlen) ((i)/(C_(i)) + (j)/(C_(j))) = (n^(2))/(2) sum_(r=0)^(n) (1)/(C_(r))` .

Explanation

Let `S=underset(0leiltjlen)(sumsum) ((i)/(C_(i)) + (j)/(C_(j)))`…(i)
Replacing I by n-I and j by n-j , we get
`S=underset(0leiltjlen)(sumsum) ((n-i)/(C_(n-i))+ (n-j)/(C_(n-j)))=underset(0leiltjlen)(sumsum) ((n-i)/(C_(i)) + (n-j)/(C_(j))) " " [because C_(r) = C_(n-r)]` ...(ii)
On adding Eqs .(i) and (ii) , we get
`2S=n underset(0leiltjlen)(sumsum) ((1)/(C_(i)) + (1)/C_(j))`
`therefore S=n/2 underset(0leiltjlen)(sumsum) ((1)/(C_(i)) + (1)/C_(j))=n/2 (sum_(r=0)^(n-1) (n-r)/(C_(r) ) + sum_(r=1)^(n) (r)/(C_(r)))`
`=n/2 (sum_(r=0)^(n-1) (n-r)/(C_(r) ) + sum_(r=1)^(n) (r)/(C_(r)))=n/2(sum_(r=0)^(n) (n)/(C_(r))) = (n^(2))/(2) sum_(r=0)^(n) (1)/(C_(r))` .