JEE Mathematics 2025 Question with Solution

Given: We need the least value of $n$ for which the number of integral terms in the binomial expansion is $183$.

Find: The least integer $n$.

From the solution working, consider the general term of

For $T_k$ to be an integer, both exponents must be integers.

So the conditions are

and

Hence,

for some integer $p$, and $n-k$ must be divisible by $3$.

The number of admissible values is therefore taken as

Given that the number of integral terms is $183$,

So,

Now solve the inequality coming from the floor function:

That is,

Therefore, the least integer value is

So the correct option is C.

Note: The solution works with $\left(7^{\frac{1}{3}} + 11^{\frac{1}{12}}\right)^n$ and concludes $n = 2184$. The final answer on the page is C, which matches option $2184$.