If `a_(n) = sum_(r=0)^(n) (1)/(""^(n)C_(r))`, then the value of `(sumsum)_(0leiltjlen) (i/(""^(n)C_(i))+j/(""^(n)C_(j)))`

Options

A. `an^(2)`
B. `(a^(2)n)/(2)`
C. `a^(2)n`
D. `(n^(2)a)/(2)`

Answer

D

Explanation

`S = underset(0leiltjlen)(sumsum)((i)/(.^(n)C_(i))+(j)/(.^(n)C_(j)))`
`= underset(0leiltjlen)(sumsum)((n-i)/(.^(n)C_(n-i))+(n-j)/(.^(n)C_(n-j)))`
`= n underset(0leiltjlen)(sumsum)((1)/(.^(n)C_(i))+(1)/(.^(n)C_(j)))-S`
`rArr S = n/2underset(0leiltjlen)(sumsum) ((1)/(.^(n)C_(i))+(1)/(.^(n)C_(j)))`
`=n/2(underset(r=0)overset(n-1)sum(n-r)/(.^(n)C_(r))+underset(r=1)overset(n)sum(r)/(.^(n)C_(r)))`
`= n/2(overset(n)underset(r=0)sum(n)/(.^(n)C_(r))) = (n^(2)a)/(2)`