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The value of `sum_(r=1)^(10) r. (""^(n)C_(r))/(""^(n)C_(r-1)` is equal toA. ` 5 (2n -9)`B. `10`nC. `9 (n-4)`D. none of these |
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Answer» Correct Answer - a We have , `sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = sum_(r=1)^(10) (n-r+1) ` `rArr sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = sum_(r=1)^(10) {(n+1) - r}` ` rArr sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = 10 (n+1) - sum_(r=1)^(10) r` `rArr sum_(r=1)^(10)r. (""^(n)C_(r))/(""^(n)C_(r-1)) = 10 (n+1) -55= 10 n -45 = 5 (2n - 9).` |
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