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Ratio of the 5^(th) term from the beginn...

Ratio of the `5^(th)` term from the beginning to the `5^(th)` term from the end in the binomial expansion of `(2^(1//3)+(1)/(2(3)^(1//3)))^(10)` is

A

`1 : 2(6)^(1)/(3)`

B

`1 : 4(16)^(1)/(3)`

C

`4 (36)^(1)/(3) : 1`

D

`2 (36)^(1)/(3) : 1`

Text Solution

Verified by Experts

Since, rth term from the end in the expansion of a binomial `(x + a)^(n)` is same as the `(n - r + 2)` th term from the beginning in the expansion of same binomial.
`:.` Raquired ratio `= (T_(6))/(T_(10 - 5 + 2)) = (T_(5))/(T_(7)) = (T_(4 + 1))/(T_(6 + 1))`
`implies (T_(6))/(T_(10 - 5 + 2)) = (.^(10)C_(4) (2^(1//3))^(10 - 4) ((1)/(2(3)^(1//3)))^(4))/(.^(10)C_(6) (2^(1//3))^(10-6) ((1)/(2(3)^(1//3)))^(6)`
`[:' T_(r + 1) = .^(n)C_(r ) x^(n - r) a^(r )]`
`= (2^(6//3) (2(3)^(1//3))^(6))/(2^(4//3)(2(3)^(1//3))^(4))" "[:' .^(10)C_(4) = .^(10)C_(6)]`
`= 2^(6//3 - 4//3) (2(3)^(1//3))^(6 - 4)`
`= 2^(23) 2^(2). 3^(23) = 4(6)^(23) = 4 36)^(1//3)`
So, the required ratio is `4(36)^(1//3) : 1`
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