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If `sum_(i = 1)^(20) ((""^(20)C_(i - 1))/(""^(20)C_(i) + ""^(20)C_(i - 1)))^(3) = k/21`, then k equalsA. 100B. 400C. 200D. 50 |
Answer» Correct Answer - A `overset(20)underset(i = 1)sum ((""^(20)C_(i - 1))/(""^(20)C_(i) + ""^(20)C_(i - 1)))^(3) = k/21` `rArr " "overset(20)underset(i = 1)sum ((""^(20)C_(i - 1))/(""^(21)C_(i)))^(3) = k/21 " "(because ""^(n)C_(r) + ""^(n)C_(r - 1) = ""^(n + 1)C_(r))` `rArr" "overset(20)underset(i = 1)sum ((""^(20)C_(i - 1))/(21/i""^(20)C_(i-1)))^(3) = k/21 " "(because ""^(n)C_(r) = n/r " "^(n -1)C_(r - 1))` `rArr" "overset(20)underset(i = 1)sum (i/21)^(3) = k/21` `rArr" "(1)/((21)^(3))overset(20)underset(i = 1)sum i^(3) = k/21` |
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