Given positive integers r 1, n 2 and that the coefficient of (3r)th and (r + 2)th terms in the b

Q: Wait, why is the index $k$ equal to $m-1$ and not $m$?

That is a fantastic question! It is the most common point of confusion in binomial theorem problems. The formula $T_{k+1} = \binom{N}{k} x^k$ defines the $(k+1)$-th term. If you want the $m$-th term, you must set $k+1 = m$, which implies $k = m-1$. Always remember: the index $k$ is the power of $x$, and it is always one less than the term position.

Q: Why do we have two cases for $\binom{N}{a} = \binom{N}{b}$?

Think about Pascal's Triangle. The row $N$ is symmetric. For example, in the 4th row (1, 4, 6, 4, 1), $\binom{4}{1} = \binom{4}{3} = 4$. Here, $a=1$ and $b=3$, and $a+b = 4 = N$. The other case, $a=b$, occurs when you are looking at the exact same term (e.g., $\binom{4}{2} = \binom{4}{2}$). Since the problem implies two distinct terms (the $(3r)$-th and $(r+2)$-th), we must consider both the possibility that they are the same term or that they are symmetric counterparts.

Q: Why did we reject $r=1$?

Excellent catch! In mathematics, we must always respect the constraints given in the problem. The question explicitly states $r > 1$. When we solved the first case ($3r-1 = r+1$), we found $r=1$. Since this violates the given condition, it is an extraneous solution and must be discarded. Always check your constraints at the end of your derivation!

Question

Given positive integers $r > 1, n > 2$ and that the coefficient of $(3r)$th and $(r + 2)$th terms in the binomial expansion of $(1 + x)^{2n}$ are equal. Then

Options:

(A) $n = 2r$

(B) $n = 2r + 1$

(C) $n = 3r$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to dive deep into the heart of the Binomial Theorem. It is not just about expanding brackets; it is about understanding the hidden symmetry of numbers.

Our problem asks us to find a relationship between $n$ and $r$ given that the coefficients of the $(3r)$-th and $(r+2)$-th terms in the expansion of $(1+x)^{2n}$ are equal.

### Decoding the Terminology

The first hurdle is the definition of the $m$-th term. We know the general term formula is $T_{k+1} = \binom{N}{k} x^k$. Here, our total power $N$ is $2n$.

A common trap is to use $k=m$ for the $m$-th term. But look closely at the formula: the term index is $k+1$. Therefore, for the $m$-th term, we must use $k = m-1$.

For the $(3r)$-th term, our index $k$ becomes $3r-1$. Thus, the coefficient is $\binom{2n}{3r-1}$.

Similarly, for the $(r+2)$-th term, our index $k$ becomes $r+1$, giving us the coefficient $\binom{2n}{r+1}$. We have successfully translated the problem into the language of binomial coefficients:

$$\binom{2n}{3r-1} = \binom{2n}{r+1}$$

### The Symmetry of Pascal's Triangle

Now, we invoke the powerful symmetry property of binomial coefficients. We know that $\binom{N}{a} = \binom{N}{b}$ if and only if $a = b$ or $a + b = N$.

This is the key that unlocks the door. We have two paths to explore:
1. The direct equality: $3r-1 = r+1$.
2. The sum of indices: $(3r-1) + (r+1) = 2n$.

### The Fork in the Road

Let us test the first path. If $3r-1 = r+1$, then $2r = 2$, which simplifies to $r = 1$.

But wait! Pause for a moment. The problem explicitly states that $r > 1$. This is a classic JEE trap—a solution that is mathematically correct in isolation but violates the problem's constraints. We must reject $r=1$.

Now, let us walk the second path. If $(3r-1) + (r+1) = 2n$, the algebra becomes incredibly satisfying. The $-1$ and $+1$ cancel out, leaving us with:

$$4r = 2n$$

Dividing both sides by $2$, we arrive at the elegant result:

**$n = 2r$**

### The Final Revelation

We have navigated the constraints and the algebra to find that **$n = 2r$**. This is not just an answer; it is a geometric statement about the placement of these terms in the binomial expansion.

They are perfectly balanced on either side of the central term. You have successfully mastered the logic of binomial coefficients. Keep this analytical mindset, and you will conquer any problem that comes your way.

Animated Solution for Mathematics - Binomial Theorem: Given positive integers $r > 1, n > 2$ and that the coefficient of $(3 r)$ th and $(r + 2)$ th terms in the binomial expansion of $(1 + x)^{2 n}$ are equal. Then

Select Answer:

Visualized Solution

Identify the General Term Formula

Coefficient of the $(3 r)$ -th Term

Coefficient of the $(r + 2)$ -th Term

Equating the Coefficients

Case 1: Direct Equality

Case 2: Sum of Indices

Final Conclusion

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

Decoding the Terminology

The Symmetry of Pascal's Triangle

The Fork in the Road

The Final Revelation

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