In the expansion of `(3^(-x//4)+3^(5x//4))^(n)` the sum of binomial coefficient is 64 and term with the greatest binomial coefficient exceeds the third by `(n-1)` , the value of `x` must be `0` b. `1` c. `2` d. `3`

Explanation

Given sum of the binomaila coefficients in the expansion of
`(3^(-x//4) + 3^(5x//4))^(n) = 64 `
Then , putting ` x^(-x//4) = 3^(5x//4) = 1`
` therefore (1 + 1)^(n) = 64 rArr 2^(n) = 2^(6)`
` therefore = 6`
We know that , middle term has the greatest binomial
coefficients . Hence , n= 6
` therefore ` Middle term `= ((n)/(2) + 1)` th term = 4 th term = ` T_(4)`
and given that ` T_(4) = (n-1) + T_(3)`
` rArr T_(3 + 1) = (6-1) + T_(2 + 1)`
` rArr ""^(6)C_(3) (3^(-x//4))^(3) (3^(5x//4))^(3) = 5 + ""^(6)C_(2) (3^(-x//4))^(4) (3^(5x//4))^(2)`
` rArr 20.3^(3x) = 5 + 15. 3^(3x//3)`
Let ` 3^(3x/2) = t `
` therefore 20 t^(2) = 5 + 15 t `
` rArr 4t^(2) - 3t - 1 = 0 `
`rArr (4t + 1) (t - 1) = 0 `
` therefore t = 1, 1 ne - (1)/(4) rArr 3 ^(3x//2) = 1= 3^(0)`
` (3x)/(2) = 0 "or : x = 0 ` .