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If `underset(i=1)overset(20)Sigma((.^(20)C_(i-1))/(.^(20)C_(i)+.^(20)C_(i-1)))^(3)=(k)/(21)`, then k equals

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The correct Answer is:
5
`(""^(20)C_(i-1)0/(""^(20)C_(i)+""^(20)C_(i-1))=(""^(20)C_(i-1))/(""^(21)C_(i))rArr((20!)/((20-i+1)!(i-1)!))/((21!)/((21-i)i!))=i/21rArrsum_(i=1)^(20)(i^(3))/(21^(3))=k/21`
`(20^(2)xx21^(2))/(4xx21^(3))=k/21`
k=100
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