Q.Given the integers $r > 1$, $n > 2$, and coefficients of $(3r)^{\text{th}}$ and $(r + 2)^{\text{nd}}$ terms in the binomial expansion of $(1 + x)^{2n}$ are equal, then
(A) $n = 2r$
(B) $n = 3r$
(C) $n = 2r + 1$
(D) none of these

The core concept here is the Combinations Symmetry Property.

The general term in the binomial expansion of $(1+x)^{2n}$ is $T_{k+1} = \binom{2n}{k} x^k$. The coefficient of the $(k+1)^{\text{th}}$ term is $\binom{2n}{k}$.

1. The coefficient of the $(3r)^{\text{th}}$ term is $\binom{2n}{3r-1}$.
2. The coefficient of the $(r+2)^{\text{nd}}$ term is $\binom{2n}{r+1}$.
3. Given that these coefficients are equal: $\binom{2n}{3r-1} = \binom{2n}{r+1}$.
4. Using the property that if $\binom{N}{A} = \binom{N}{B}$, then either $A=B$ or $A+B=N$: …

When dealing with binomial expansions, understanding the structure of terms and their coefficients is fundamental. The expression $(1+x)^{2n}$ is a binomial raised to the power $2n$.

Let's first recall the general term in a binomial expansion.

For an expansion of $(a+b)^N$, the $(k+1)^{\text{th}}$ term is given by $T_{k+1} = {}^N C_k a^{N-k} b^k$.

In our case, the binomial is $(1+x)^{2n}$. Here, $a=1$, $b=x$, and the total power is $N=2n$.

So, the $(k+1)^{\text{th}}$ term is $T_{k+1} = {}^{2n} C_k (1)^{2n-k} x^k = {}^{2n} C_k x^k$.

The coefficient of the $(k+1)^{\text{th}}$ term is simply ${}^{2n} C_k$.

The core concept for this problem is the symmetry property of binomial coefficients.

Let's apply this understanding to the given problem step-by-step.

1. Identify the coefficients of the specified terms.

   • The $(3r)^{\text{th}}$ term: For this term, $k+1 = 3r$, which means $k = 3r-1$. The coefficient of the $(3r)^{\text{th}}$ term is ${}^{2n} C_{3r-1}$.
   • The $(r+2)^{\text{nd}}$ term: For this term, $k+1 = r+2$, which means $k = r+1$. The coefficient of the $(r+2)^{\text{nd}}$ term is ${}^{2n} C_{r+1}$.

2. Set the coefficients equal as per the problem statement.

   We are given that these two coefficients are equal:

   ${}^{2n} C_{3r-1} = {}^{2n} C_{r+1}$

3. Apply the symmetry property of binomial coefficients.

   Here, $N = 2n$, $A = 3r-1$, and $B = r+1$.

   According to the property ${}^N C_A = {}^N C_B \implies A=B$ or $A+B=N$, we have two cases:

   • Case 1: The lower indices are equal. $3r-1 = r+1$ $3r - r = 1 + 1$ $2r = 2$ $r = 1$ …