JEE Mathematics 2025 Question with Solution

Given: The expansion is $\left( \sqrt{5} + \frac{1}{\sqrt{5}} \right)^n$ and the ratio of the $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$.

Find: The value of $^nC_3$.

Using the binomial theorem, the general term is

$$T_{r+1} = {}^nC_r \left( \sqrt{5} \right)^{n-r} \left( \frac{1}{\sqrt{5}} \right)^r = {}^nC_r \left( \sqrt{5} \right)^{n-2r}$$

The $15^{th}$ term from the beginning is

$$T_{15} = {}^nC_{14} \left( \sqrt{5} \right)^{n-28}$$

The $15^{th}$ term from the end is the $\left(n-13\right)^{th}$ term from the beginning, so

$$T_{n-13} = {}^nC_{14} \left( \sqrt{5} \right)^{28-n}$$

Now,

$$\frac{T_{15}}{T_{n-13}} = \frac{{}^nC_{14} \left( \sqrt{5} \right)^{n-28}}{{}^nC_{14} \left( \sqrt{5} \right)^{28-n}} = \left( \sqrt{5} \right)^{2n-56}$$

So,

$$\left( \sqrt{5} \right)^{2n-56} = \frac{1}{6}$$

From the extracted the solution, the worked result leads to $n = 25$, and hence

$${}^nC_3 = {}^{25}C_3 = \frac{25 \cdot 24 \cdot 23}{3 \cdot 2 \cdot 1} = 2300$$

Therefore, the correct option is C.

Given: We need the ratio of the $15^{th}$ term from the beginning to the $15^{th}$ term from the end in the expansion.

Find: $\binom{n}{3}$.

The solution states the term formula

$$T_{r+1}=\binom{n}{r} a^{n-r} b^r$$

Here,

$$a=\sqrt[3]{2}, \qquad b=\frac{1}{\sqrt[3]{3}}$$

were used in the alternate approach shown on the page, and it computes:

$$T_{\text{beg}}=\binom{n}{14}\,2^{\frac{n-14}{3}}\,3^{-\frac{14}{3}}$$ $$T_{\text{end}}=\binom{n}{14}\,2^{\frac{14}{3}}\,3^{-\frac{n-14}{3}}$$

Thus,

$$\frac{T_{\text{beg}}}{T_{\text{end}}}=6^{\frac{n-28}{3}}$$

Given that this ratio is $\frac{1}{6}=6^{-1}$,

$$\frac{n-28}{3}=-1$$ $$n=25$$

Now evaluate

$$\binom{25}{3}=\frac{25\cdot24\cdot23}{3\cdot2\cdot1}=2300$$

Therefore, the required value is $2300$ and the correct option is C.

Note: The solution contains inconsistent intermediate expressions across its two approaches, but both conclude with the same final answer $2300$, matching option C.