For any n `in` N, let `C_(r)` stand for `""^(n)C_(r)`,
` r = 0,1,2,3,…,n` and let `S= sum_(r=0)^(n) (1)/(C_(r))`
Statement-1: `underset(0leilt i le n)(sumsum) ((i)/(C_(i))+(j)/(C_(j)))= (n^(n))/(2)S`
Statement-2:` underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nS`

Options

A. 1
B. 2
C. 3
D. 4

Answer

a

Explanation

We have,
`underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= sum_(i=0)^(n-1) (n-i)/(C_(i) )+ sum_(j=1)^(n) (j)/(C_(j))`
`rArr underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= n sum_(i=0)^(n-1) (1)/(C_(i) )- sum_(i=0)^(n) (i)/(C_(i)) + sum_(j=1)^(1)(j)/(C_(j))`
`rArr underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nsum_(i=0)^(n-1) (1)/(C_(i) )+(n)/(C_(n))`
`rArr underset(0leilt i le n)(sumsum) ((1)/(C_(i))+(1)/(C_(j)))= nsum_(i=0)^(n-1) (1)/(C_(i) )=nS`
So, statement-2 is true.
Let `S_(1) underset(0leiltj len) (sumsum) ((i)/(C_(i))+(j)/(C_(j)))` . Then
`S_(1) underset(0leiltj len) (sumsum) ((n-i)/(C_(n-i))+(n-j)/(C_(n-j)))` .
`rArr S_(1) = n underset(0leiltj len) (sumsum) ((1)/(C_(n-i))+(1)/(C_(n-j)))- underset(0leiltj len) (sumsum)((i)/(C_(n-i))+(j)/(C_(n-j)))`
`rArr S_(1) = n underset(0leiltj len) (sumsum) ((1)/(C_(i))+(1)/(C_(j)))- underset(0leiltj len) (sumsum)((i)/(C_(i))+(j)/(C_(j)))`
`rArr S_(1) = n(nS) - S_(1)` [ Using truth of statement-2]
`rArr 2S_(1) = n^(2) S rArr S_(1) = (n^(2))/(2)S`
So, statement-1 is true and stetement-2 is a correct expanation
for statement-1.