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If underset(i=1)overset(20)Sigma((.^(20)...

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

A

100

B

400

C

200

D

50

Text Solution

Verified by Experts

The correct Answer is:
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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