Finding the Middle Term Coefficient in Binomial Expansion

We are asked to find the coefficient of the middle term in the expansion of (1 + 4x + 4x^2)^5. To simplify this, let's first look at the expression inside the parenthesis: 1 + 4x + 4x^2. This is a perfect square trinomial, which can be factored.

Recognizing the pattern a^2 + 2ab + b^2 = (a+b)^2:

• Here, a^2 = 1, so a = 1.
• Also, b^2 = 4x^2, so b = \sqrt{4x^2} = 2x.
• Let's check the middle term: 2ab = 2(1)(2x) = 4x. This matches the given expression.

So, 1 + 4x + 4x^2 = (1 + 2x)^2.

Now, the original expression becomes ((1 + 2x)^2)^5. Using the exponent rule (a^m)^n = a^{mn}, we get:

( (1 + 2x)^2 )^5 = (1 + 2x)^{2 \times 5} = (1 + 2x)^{10}.

Determining the Middle Term Position

We need to find the middle term in the expansion of (1 + 2x)^{10}. For a binomial expansion of the form (a+b)^n, the total number of terms is n+1.

In our case, n = 10. So, the number of terms is 10 + 1 = 11.

Since the total number of terms (11) is odd, there is a single middle term. The position of the middle term is given by the formula \frac{n}{2} + 1 for even n.

For n = 10, the position of the middle term is \frac{10}{2} + 1 = 5 + 1 = 6^{th} term.

Calculating the Middle Term

The general term, T_{r+1}, in the binomial expansion of (a+b)^n is given by the formula:

T_{r+1} = \binom{n}{r} a^{n-r} b^r

For the 6^{th} term, we have r+1 = 6, which means r = 5.

In the expansion of (1 + 2x)^{10}:

• n = 10
• a = 1
• b = 2x
• r = 5

Substitute these values into the general term formula:

T_6 = T_{5+1} = \binom{10}{5} (1)^{10-5} (2x)^5

T_6 = \binom{10}{5} (1)^5 (2^5 x^5)

T_6 = \binom{10}{5} \cdot 1 \cdot 32x^5

T_6 = \binom{10}{5} \cdot 32 \cdot x^5

Calculating the Binomial Coefficient

Now we need to calculate the binomial coefficient \binom{10}{5}:

\binom{n}{r} = \frac{n!}{r! (n-r)!}

\binom{10}{5} = \frac{10!}{5! (10-5)!} = \frac{10!}{5! 5!}

\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1) \times (5 \times 4 \times 3 \times 2 \times 1)}

We can cancel out 5! from the numerator and denominator:

\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}

Calculate the value:

\binom{10}{5} = \frac{30240}{120}

\binom{10}{5} = 252

Finding the Coefficient

The middle term was found to be \binom{10}{5} \cdot 32 \cdot x^5. The coefficient of this term is the part that does not include x, which is \binom{10}{5} \cdot 32.

Coefficient = 252 \times 32

Let's perform the multiplication:

So, the coefficient of the middle term is 8064.

Conclusion

The expansion of (1 + 4x + 4x^2)^5 is equivalent to the expansion of (1 + 2x)^{10}. The middle term in this expansion is the 6^{th} term, and its coefficient is calculated as \binom{10}{5} \times 2^5, which equals 252 \times 32 = 8064.

Revision Table: Key Steps

Additional Information: Binomial Theorem Concepts

The Binomial Theorem provides a formula for expanding expressions of the form (a+b)^n for any positive integer n.

The expansion is given by:

(a+b)^n = \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}{r}a^{n-r} b^r + \dots + \binom{n}{n}a^0 b^n

Key aspects:

• Each term has the form \binom{n}{r}a^{n-r} b^r, where r goes from 0 to n.
• The sum of the powers of a and b in each term is always n ((n-r) + r = n).
• The coefficients \binom{n}{r} are called binomial coefficients and can be calculated using factorials or found from Pascal's Triangle.
• The term with b^r is the (r+1)^{th} term in the expansion.

When n is even, there is one middle term at position \frac{n}{2} + 1. When n is odd, there are two middle terms at positions \frac{n+1}{2} and \frac{n+1}{2} + 1.