Question

In the expansion of $ (1 + 3x + 3x^2 + x^3)^{2n} $, the term which has the greatest binomial coefficient, is

• A. \( (3n) \, \text{th term} \)
• B. \( (3n + 1) \, \text{th term} \)
• C. \( (3n - 1) \, \text{th term} \)
• D. \( (3n + 2) \, \text{th term} \)

Hint:

In multinomial expansions, the binomial coefficient is largest at the middle terms. For even degree expansions, the middle term is the one with the largest coefficient.

Answer 1

The Correct Option is B

Problem Analysis:
We need to determine which term is the middle term in the expansion of \((1 + 3x + 3x^2 + x^3)^{2n}\).

1. Expression Simplification:
First, recognize that the given expression can be rewritten as a perfect cube: \[ 1 + 3x + 3x^2 + x^3 = (1 + x)^3 \] Therefore, the original expression becomes: \[ (1 + x)^{6n} \]

2. Expansion Properties:
The expansion of \((1 + x)^{6n}\) has: \[ 6n + 1 \text{ terms} \] Since \(6n\) is always even, there is exactly one middle term.

3. Middle Term Identification:
For an expansion with \(N + 1\) terms where \(N\) is even: \[ \text{Middle term position} = \frac{N}{2} + 1 \] Applying this to our case: \[ \frac{6n}{2} + 1 = 3n + 1 \]

4. Verification:
- For \(n=1\) (expansion of \((1+x)^6\)): middle term is 4th term (\(3(1)+1=4\)) - For \(n=2\) (expansion of \((1+x)^{12}\)): middle term is 7th term (\(3(2)+1=7\)) Both cases confirm our general formula.

Final Answer:
The middle term in the expansion is the \((3n + 1)\)-th term.

Answer 2

To find the term with the greatest binomial coefficient in the expansion of \( (1 + 3x + 3x^2 + x^3)^{2n} \), we must identify the point where the product of binomial coefficients reaches its peak.

The expression can be seen as a multinomial expansion, which involves terms of the form: \(C \cdot (1)^a \cdot (3x)^b \cdot (3x^2)^c \cdot (x^3)^d\) where \(a + b + c + d = 2n\). Here, \(C\) is the multinomial coefficient:

\(\frac{(2n)!}{a!b!c!d!}\)

The general term corresponding to the sequence \(a, b, c, d\) is:

\(\frac{(2n)!}{a!b!c!d!} \cdot 3^b \cdot (3^2)^c \cdot (x^{b+2c+3d})\)

We maximize \(\frac{(2n)!}{a!b!c!d!}\) under the constraint \(a + b + c + d = 2n\).

For the symmetric distribution of terms, setting the exponents approximately equal will help find the term with maximum coefficient. Assume equal distribution, \(b+2c+3d = 3n\), then solving for a balanced split:

Variable	Value
a	\(n\)
b	\(n/2\)
c	\(n/2\)
d	\(n/2\)

In this configuration, \(b + 2c + 3d = 3n/2\), which aligns with the central term calculations.

For large \(n\), the multinomial coefficients have a term maximized near \( (3n+1)\), as it aligns the central exponent distribution for \(3x\), \(3x^2\), \(x^3\).

Therefore, the term with the maximum binomial coefficient is the \((3n+1)^{th}\) term. Hence:

The correct answer is: \( (3n + 1) \, \text{th term} \)