Key Concept: Binomial Theorem and General Term

The binomial theorem provides a formula for the algebraic expansion of powers of a binomial $(a+b)^N$. For the expansion of $(1+x)^N$, the general term, often denoted as the $(k+1)^{th}$ term, is given by: $T_{k+1} = {^N C_k} (1)^k (x)^k = {^N C_k} x^k$ where $^N C_k = \frac{N!}{k!(N-k)!}$ is the binomial coefficient. In this specific problem, the expansion is of $(1+x)^{2n}$, so $N = 2n$. Therefore, the general term is $T_{k+1} = {^{2n}C_k} x^k$. The coefficient of this term is ${^{2n}C_k}$.

Step-by-Step Working

1. Identifying the Coefficient of the $(r+2)^{th}$ Term

• We need to find the coefficient of the $(r+2)^{th}$ term in the expansion of $(1+x)^{2n}$.
• Using the general term formula $T_{k+1} = {^{2n}C_k} x^k$, we compare the term number: $k+1 = r+2$.
• Subtracting 1 from both sides, we find the value of $k$: $k = (r+2) - 1 = r+1$.
• Substituting this value of $k$ back into the general term, the $(r+2)^{th}$ term is $T_{r+2} = {^{2n}C_{r+1}} x^{r+1}$.
• The coefficient of the $(r+2)^{th}$ term is thus ${^{2n}C_{r+1}}$.

2. Identifying the Coefficient of the $3r^{th}$ Term

• Similarly, we need to find the coefficient of the $3r^{th}$ term in the expansion of $(1+x)^{2n}$.
• Comparing the term number with $k+1$, we have $k+1 = 3r$.
• Solving for $k$: $k = 3r - 1$.
• Substituting this $k$ into the general term, the $3r^{th}$ term is $T_{3r} = {^{2n}C_{3r-1}} x^{3r-1}$.
• The coefficient of the $3r^{th}$ term is ${^{2n}C_{3r-1}}$.

3. Applying the Given Condition

• The problem states that the coefficient of the $(r+2)^{th}$ term and the $3r^{th}$ term are equal.
• Equating the coefficients we found: ${^{2n}C_{r+1}} = {^{2n}C_{3r-1}}$

4. Using Properties of Binomial Coefficients

• A crucial property of binomial coefficients states: If ${^N C_p} = {^N C_q}$, then there are two possibilities:

  1. $p = q$
  2. $p+q = N$

• In our equation, $N=2n$, $p=r+1$, and $q=3r-1$.

• Case 1: $p = q$

  • $r+1 = 3r-1$
  • To solve for $r$, we rearrange the terms: $1+1 = 3r-r \Rightarrow 2 = 2r \Rightarrow r = 1$.
  • However, the problem statement explicitly mentions that $r > 1$. Therefore, this case where $r=1$ is not valid under the given constraints.

• Case 2: $p+q = N$

  • Substituting our values: $(r+1) + (3r-1) = 2n$.
  • Combine like terms: $r + 3r + 1 - 1 = 2n$.
  • This simplifies to: $4r = 2n$.
  • Dividing both sides by 2, we get: $n = 2r$.

5. Verification of Constraints

• We are given that $r > 1$ and $n > 2$.
• Our solution $n = 2r$ is consistent with these conditions. For example, if we take the smallest integer value for $r$ such that $r > 1$ (i.e., $r=2$), then $n = 2(2) = 4$. Since $4 > 2$, the solution holds for valid values of $r$.

Tips and Common Mistakes to Avoid

• Index Confusion: A very common mistake is confusing the term number ($T_m$) with the index ($k$) in ${^N C_k}$. Always remember that the $m^{th}$ term corresponds to $k=m-1$.
• Ignoring Constraints: Always check the given conditions ($r>1, n>2$) at each stage. They are often there to eliminate extraneous solutions or validate your findings.
• Forgetting Properties: Make sure to recall all relevant properties of combinations. Missing one part of the $^N C_p = ^N C_q$ property (i.e., $p+q=N$) would lead to an incorrect conclusion if the other part ($p=q$) is invalid.

Summary/Key Takeaway

This problem is a straightforward application of the binomial theorem's general term and a fundamental property of binomial coefficients. When two binomial coefficients with the same upper index are equal ($^N C_p = ^N C_q$), the lower indices ($p$ and $q$) are either equal or their sum equals the upper index ($p+q=N$). By carefully considering the problem's constraints, we can determine the correct relationship between $n$ and $r$. The solution demonstrates that $n=2r$ is the only valid relationship given $r>1$.