InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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`A. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - a 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. |
|