r and n are positive integers r 1, n 2 and coefficient of (r+2)th term and 3rth term in the expa

Q: Wait, why is the index $k$ equal to $r+1$ for the $(r+2)^{\text{th}}$ term? Shouldn't it be $r+2$?

Great question! Remember that the general term is $T_{k+1} = \binom{N}{k}x^k$. The term number is always one greater than the index $k$. So, for the $(r+2)^{\text{th}}$ term, we set $k+1 = r+2$, which gives $k = r+1$. It's a classic trap, but once you see the pattern, you'll never fall for it again!

Q: Why do we have to consider two cases for $\binom{N}{x} = \binom{N}{y}$?

Because the binomial distribution is symmetric! The coefficients increase to a peak and then decrease. If two coefficients are equal, they are either the same term (the trivial case $x=y$) or they are two different terms equidistant from the center (the symmetric case $x+y=N$). We must check both to be mathematically rigorous.

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

The problem explicitly states $r > 1$. If we had accepted $r=1$, we would be looking at the same term twice, which doesn't fit the context of comparing two distinct terms in the expansion. Always check your constraints at the end of the problem!

Question

$r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r+2)^{th}$ term and $3r^{th}$ term in the expansion of $(1 + x)^{2n}$ are equal, then $n$ equals

Options:

(A) $3r$

(B) $3r + 1$

(C) $2r$

(D) $2r + 1$

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Symmetry

Welcome, fellow explorer of the mathematical universe! Today, we are diving into the elegant world of binomial expansions.

Imagine the expansion of $(1+x)^{2n}$. It is not just a string of numbers; it is a beautiful, symmetric mountain of coefficients.

Our mission is to find the relationship between $n$ and $r$ given that the $(r+2)^{	ext{th}}$ and $3r^{	ext{th}}$ terms have equal coefficients. Let us embark on this journey together.

### The General Term: The DNA of the Expansion

Every binomial expansion has a heartbeat, and that is the general term. For any expansion of the form $(1+x)^N$, the $(k+1)^{	ext{th}}$ term is given by:

$$T_{k+1} = \binom{N}{k}x^k$$

In our specific case, the exponent is $2n$, so our general term is $T_{k+1} = \binom{2n}{k}x^k$. The coefficient of this term is simply $\binom{2n}{k}$.

Think of this as the DNA of the expansion—it tells us everything we need to know about any specific term we might encounter.

### Translating the Problem into Math

Now, let us translate the problem's requirements into our mathematical language. We are told that the coefficient of the $(r+2)^{	ext{th}}$ term is equal to the coefficient of the $3r^{	ext{th}}$ term.

To find the coefficient of the $(r+2)^{	ext{th}}$ term, we set the term index $k+1 = r+2$, which gives us $k = r+1$. Thus, the coefficient is $\binom{2n}{r+1}$.

Similarly, for the $3r^{	ext{th}}$ term, we set $k+1 = 3r$, which gives us $k = 3r-1$. The coefficient is $\binom{2n}{3r-1}$.

Equating these, we get the fundamental equation:

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

### The Symmetry Property

Here is where the magic happens. We know that for any binomial coefficient, $\binom{N}{x} = \binom{N}{y}$ implies one of two things: either $x = y$ (the trivial case) or $x + y = N$ (the symmetric case).

This symmetry is the heartbeat of Pascal's triangle. We must analyze both possibilities to ensure we do not miss any solutions.

### Solving the Algebraic Dance

**Case 1: $x = y$**

This leads to $r+1 = 3r-1$. Rearranging this, we get $2r = 2$, which means $r = 1$.

However, the problem explicitly states that $r > 1$. Therefore, we must reject this case. It is a vital lesson: always check your constraints!

**Case 2: $x + y = N$**

This is the symmetric case. Here, the sum of the indices must equal the total power:

$$(r+1) + (3r-1) = 2n$$

Simplifying the left side, we get $4r = 2n$. Dividing both sides by $2$, we find the elegant result:

$$n = 2r$$

### Conclusion

We have successfully navigated the problem! The relationship is **$n = 2r$**.

This journey through binomial coefficients shows us that math is not just about calculation; it is about recognizing patterns and understanding the underlying symmetry of the world. Keep practicing, stay curious, and remember that every problem is just a puzzle waiting to be solved.

Animated Solution for Mathematics - Binomial Theorem: $r$ and $n$ are positive integers $r > 1, n > 2$ and coefficient of $(r + 2)^{t h}$ term and $3 r^{t h}$ term in the expansion of $(1 + x)^{2 n}$ are equal, then $n$ equals

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Visualized Solution

General Term of $(1 + x)^{2 n}$

Coefficient of $(r + 2)^{th}$ Term

Coefficient of $3 r^{th}$ Term

Equating the Coefficients

Property of $(x N) = (y N)$

Case 1: $x = y$

Evaluating Case 1

Case 2: $x + y = 2 n$

Solving for $n$

The Way Forward

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Symmetry

The General Term

Translating the Problem into Math

The Symmetry Property

Solving the Algebraic Dance

Conclusion

Similar Questions

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Visualized Solution

General Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1+x)2n

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (r+2)th Term

Coefficient of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3rth Term

Equating the Coefficients

Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (xN​)=(yN​)

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=y