

InterviewSolution
Saved Bookmarks
1. |
Prove that `sum_(r=0)^(n) r(n-r)C_(r)^(2)=n^(2)(""^(2n-2)C_(n))`. |
Answer» `underset(r=0)overset(n)sumr(n-r)(.^(n)C_(r))^(2) = underset(r=0)overset(n)sumr.^(n)C_(r)(n-r).^(n)C_(n-r)` `=underset(r=0)overset(n)sumn.^(n-)C_(r-1)n.^(n-1)C_(n-r-1)` `= n^(2)underset(r=0)overset(n)sum.^(n-1)C_(r-1).^(n-1)C_(n-r-1)` `= n^(2) xx "coefficient of" x^(n-2) "in" (1+x)^(n-1)(1+x)^(n-1)` `= n^(2) xx .^(2n-2)C_(n-2) = n^(2).^(2n-2)C_(n)` |
|