The number of irrational terms in expansion `(7^(1/5)-3^(1/10))^60` is

Options

A. 49
B. 48
C. 54
D. 55

Answer

C

Explanation

The general term in the binomial expansionof `(a + b)^(n)` is `T_(r + 1) = .^(n)C_(r ) a^(n - r) b^(r )`
So, the general term in the binomial expansion of `(7^(1//5) - 3^(1//10))^(60)` is
`T_(r + 1) = .^(60)C_(r ) (7^(1//5))^(60 - r) (-3^(1//10))^(r )`
`= .^(60)C_(r ) 7^((60 - r)/(5)) (-1)^(r ) 3^((r )/(10)) = (-3)^(r) .^(60)C_(r ) 7^(12 - (r )/(5) 3^((r )/(10))`
The possible non-negative integral values of 'r' for which `(r )/(5)` and `(r )/(10)` are integer, where `r le 60`, are `r = 0, 10, 20, 30, 40, 50, 60`
`:.` There are 7 rational terms in the binomial expansion and remaining `61 - 7 = 54` terms are irrational terms.