InterviewSolution
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InterviewSolution
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The value of `sum_(r=0)^(n) r(n -r) (""^(n)C_(r))^(2)` is equal toA. `n^(2) ""^(2n -1)C_(n -1)`B. `n^(2) """^(2n-2)C_(n)`C. `n^(2) ""^(2n)C_(n -1)`D. `n^(2) ""^(2n -1)C_(n)` |
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Answer» Correct Answer - b `sum_(r=0)^(n) (n-r)(""^(n)C_(r))^(2)` `= n sun_(r=0)^(n) (r ""^(n)C_(r)) (""^(n)C_(r)) - sum_(r=1)^(n) (r ""^(n)C_(r))^(2)` `= n^(2) (sum_(r=1)^(n) ""^(n-1)C_(r-1) ""^(n)C_(r)) - n^(2) {sum_(r=1)^(n) (""^(n-1)C_(r-1))^(2)}` `= n^(2)` { Coeff. of` "" x^(n-1)` in` (1 + x )^(2n-1)}` ` - n^(2) `{Coeff. of `x^(n-1)` in`(1 + x)^(2n-2)} ` `= n^(2) ""^(2n-1)c_(n-1) - n^(2) ""^(2n -2)C_(n-1)` `n^(2) {((2n -1)1)/(n!(n-1)!) - ((2n -2)!)/((n-1)!(n-1)!) -n^(2) ((2n -2)!)/((n-2)!n)` ` n^(2) ""^(2n-2)C_(n)` |
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