If \(n\) is the number or irrational terms in the expansion of \(\left(3^{1 / 4}+5^{1 / 8}\right)^{60}\), then ( \(n-1\) ) is divisible by:

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Binomial Theorem

n

is the number or irrational terms in the expansion of

(3^{1/4} + 5^{1/8})^{60}

, then (

n - 1

) is divisible by:

Answer: (D)

The general term in the binomial expansion is

(r 60) (3^{1/4})^{60 - r} (5^{1/8})^{r} = (r 60) 3^{(60 - r) /4} 5^{r /8}

. For the term to be rational, both exponents must be integers:

\frac{60 - r}{4}

and

\frac{r}{8}

must be integers. This requires

r \equiv 0 (mod 8)

and

60 - r \equiv 0 (mod 4)

, i.e.,

r \equiv 0 (mod 4)

. Combined:

r \equiv 0 (mod 8)

. So

r \in {0, 8, 16, 24, 32, 40, 48, 56}

— 8 rational terms. Total terms = 61, so irrational terms

n = 61 - 8 = 53

. Then

n - 1 = 52

, which is divisible by 26.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option