Algebra – Binomial Theorem – JEE Main 29 Jan 2025 Shift 1

Solution:

The general term in the expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^{n}$ is:

$T_{r+1} = {}^{n}C_{r} (7^{1/3})^{n-r} (11^{1/12})^{r}$

$T_{r+1} = {}^{n}C_{r} \cdot 7^{\frac{n-r}{3}} \cdot 11^{\frac{r}{12}}$

For the term to be integral, the powers of 7 and 11 must be integers.

1. $\frac{r}{12}$ must be an integer $\Rightarrow r$ must be a multiple of 12.

2. $\frac{n-r}{3}$ must be an integer.

Let $r \in \{0, 1, 2, \dots, n\}$.

Since $r$ must be a multiple of 12, let $r = 12k$ where $k$ is a whole number ($k \in W$).

Also, since $r$ is a multiple of 12, $r$ is automatically divisible by 3.

For $\frac{n-r}{3}$ to be an integer, $n$ must be divisible by 3 (since $r$ is divisible by 3).

The values of $r$ are $0, 12, 24, \dots, 12(N-1)$.

Given that the number of integral terms is 183.

So, the sequence of valid $r$ values has 183 terms.

$r$ values: $12(0), 12(1), 12(2), \dots, 12(182)$.

The maximum value of $r$ is $12 \times 182 = 2184$.

For the sequence to exist up to this term, $n$ must be at least this maximum $r$.

Thus, the least value of $n = 2184$.

Ans. (1)