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If `(.^(20)C_(1))/(1)+(.^(20)C_(3))/(2)+(.^(20)C_(5))/(3)+.......+(.^(20)C_(19))/(10)=(K(2^(20-1)))/(21)`, then K is equal to : |
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Answer» Correct Answer - 2 `(.^(20)C_(1))/(1)+(.^(20)C_(3))/(2)+(.^(20)C_(5))/(3)+......+(.^(20)C_(19))/(10)` `=underset(r=1)overset(10)Sigma(1)/(r).^(20)C_(2r-1)=(2)/(21)underset(r=1)overset(10)Sigma(21)/(2r).^(20)C_(2r-1)` `=(2)/(21)underset(r=1)overset(10)Sigma.^(21)C_(2r)=(2)/(21)[.^(21)C_(2)+.^(21)C_(4)+.^(21)C_(20)]` `=(2)/(21)(2^(20)-1)` |
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