Understanding the Problem: Finding the Middle Term Coefficient

We are asked to find the coefficient of the middle term in the binomial expansion of (2 + 3x)⁴. This requires understanding the binomial theorem and how to identify the middle term in an expansion.

Binomial Expansion Basics

The binomial theorem states that the expansion of (a + b)ⁿ is given by:

\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n} a^0 b^n\)

The expansion of (a + b)ⁿ has (n + 1) terms.

In our case, the binomial is (2 + 3x)⁴, so:

• a = 2
• b = 3x
• n = 4

The number of terms in the expansion will be n + 1 = 4 + 1 = 5 terms.

Identifying the Middle Term

When the power 'n' is even, there is exactly one middle term. The position of the middle term is given by \(\left(\frac{n}{2} + 1\right)\).

For (2 + 3x)⁴, n = 4 (which is even). The position of the middle term is:

\(\text{Position} = \frac{4}{2} + 1 = 2 + 1 = 3\)

So, the 3rd term in the expansion is the middle term.

Calculating the Middle Term

The general term in the binomial expansion \( (a+b)^n \) is given by \( T_{k+1} = \binom{n}{k} a^{n-k} b^k \). The 3rd term corresponds to k + 1 = 3, which means k = 2.

Using the values a = 2, b = 3x, n = 4, and k = 2, the 3rd term \( T_3 \) is:

\(T_3 = \binom{4}{2} (2)^{4-2} (3x)^2\)

Let's calculate the components:

• \(\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6\)
• \((2)^{4-2} = (2)^2 = 4\)
• \((3x)^2 = 3^2 \times x^2 = 9x^2\)

Now, substitute these values back into the term formula:

\(T_3 = 6 \times 4 \times 9x^2\)

\(T_3 = 24 \times 9x^2\)

\(T_3 = 216x^2\)

Identifying the Coefficient

The middle term is \(216x^2\). The coefficient is the numerical part multiplying the variable term.

The coefficient of the middle term \(216x^2\) is 216.

Verification: Full Expansion (Optional)

Let's quickly look at all terms to confirm the middle one:

\((2 + 3x)^4 = \binom{4}{0} 2^4 (3x)^0 + \binom{4}{1} 2^3 (3x)^1 + \binom{4}{2} 2^2 (3x)^2 + \binom{4}{3} 2^1 (3x)^3 + \binom{4}{4} 2^0 (3x)^4\)

• \(T_1 = \binom{4}{0} 2^4 (3x)^0 = 1 \times 16 \times 1 = 16\)
• \(T_2 = \binom{4}{1} 2^3 (3x)^1 = 4 \times 8 \times 3x = 96x\)
• \(T_3 = \binom{4}{2} 2^2 (3x)^2 = 6 \times 4 \times 9x^2 = 216x^2\)
• \(T_4 = \binom{4}{3} 2^1 (3x)^3 = 4 \times 2 \times 27x^3 = 216x^3\)
• \(T_5 = \binom{4}{4} 2^0 (3x)^4 = 1 \times 1 \times 81x^4 = 81x^4\)

The terms are 16, \(96x\), \(216x^2\), \(216x^3\), \(81x^4\). With 5 terms, the 3rd term (\(216x^2\)) is indeed the middle term, and its coefficient is 216.

Conclusion: The Coefficient of the Middle Term

The middle term in the expansion of (2 + 3x)⁴ is the 3rd term, which is \(216x^2\). The coefficient of this term is 216.

Revision Table: Binomial Expansion Concepts

Additional Information: Combinations and Binomial Coefficients

The binomial coefficients \(\binom{n}{k}\) are also known as combinations, representing the number of ways to choose k items from a set of n items without regard to the order. The formula is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). These coefficients appear in Pascal's triangle, where each number is the sum of the two numbers directly above it.

For n=4, the binomial coefficients are:

• \(\binom{4}{0} = 1\)
• \(\binom{4}{1} = 4\)
• \(\binom{4}{2} = 6\)
• \(\binom{4}{3} = 4\)
• \(\binom{4}{4} = 1\)

These coefficients form the 5th row of Pascal's triangle (starting with row 0). Understanding combinations helps in calculating the terms of a binomial expansion efficiently.

The binomial theorem is fundamental in various areas of mathematics, including probability, statistics, and calculus.