Understanding Binomial Expansion Terms

The problem asks us to find the value of 'n' in the binomial expansion of (1 + ax)ⁿ, given the first three terms are 1, 12x, and 64x².

Let's recall the binomial expansion formula for (1 + y)ⁿ:

$$ (1+y)^n = \binom{n}{0} + \binom{n}{1}y + \binom{n}{2}y^2 + \binom{n}{3}y^3 + \dots + \binom{n}{n}y^n $$

In our case, y = ax. Substituting this into the formula, the expansion of (1 + ax)ⁿ is:

$$ (1+ax)^n = \binom{n}{0} + \binom{n}{1}(ax) + \binom{n}{2}(ax)^2 + \binom{n}{3}(ax)^3 + \dots $$

Let's calculate the first three terms of this expansion:

• First Term: The first term is when the power of (ax) is 0.

  $$ \binom{n}{0}(ax)^0 $$

  We know that $\binom{n}{0} = 1$ and $(ax)^0 = 1$. So, the first term is $1 \times 1 = 1$. This matches the given first term.
• Second Term: The second term is when the power of (ax) is 1.

  $$ \binom{n}{1}(ax)^1 $$

  We know that $\binom{n}{1} = n$. So, the second term is $n(ax) = nax$. According to the problem, the second term is 12x. Thus, we have our first equation:

  $$ nax = 12x $$

  Since this must hold for any x (except possibly x=0, but we are comparing coefficients), we can equate the coefficients of x:

  $$ na = 12 \quad \text{(Equation 1)} $$

• Third Term: The third term is when the power of (ax) is 2.

  $$ \binom{n}{2}(ax)^2 $$

  We know that $\binom{n}{2} = \frac{n(n-1)}{2 \times 1} = \frac{n(n-1)}{2}$. Also, $(ax)^2 = a^2x^2$. So, the third term is $\frac{n(n-1)}{2} a^2x^2 = \frac{n(n-1)a^2}{2}x^2$. According to the problem, the third term is 64x². Thus, we have our second equation by equating the coefficients of x²:

  $$ \frac{n(n-1)a^2}{2} = 64 \quad \text{(Equation 2)} $$

Solving Equations to Find n

Now we have a system of two equations with two variables, 'n' and 'a':

1. $$ na = 12 $$
2. $$ \frac{n(n-1)a^2}{2} = 64 $$

From Equation 1, we can express 'a' in terms of 'n' (assuming n ≠ 0, which is true as we have multiple terms in the expansion):

$$ a = \frac{12}{n} $$

Now substitute this expression for 'a' into Equation 2:

$$ \frac{n(n-1)}{2} \left(\frac{12}{n}\right)^2 = 64 $$

Simplify the equation:

$$ \frac{n(n-1)}{2} \times \frac{144}{n^2} = 64 $$

We can cancel one 'n' from the numerator with one 'n' from the denominator:

$$ \frac{(n-1)}{2} \times \frac{144}{n} = 64 $$

Multiply both sides by 2:

$$ (n-1) \times \frac{144}{n} = 128 $$

Multiply both sides by n:

$$ 144(n-1) = 128n $$

Distribute 144 on the left side:

$$ 144n - 144 = 128n $$

Rearrange the terms to isolate 'n':

$$ 144n - 128n = 144 $$

$$ 16n = 144 $$

Divide by 16:

$$ n = \frac{144}{16} $$

$$ n = 9 $$

Let's verify this value of n. If n = 9, then from na = 12, we get 9a = 12, so a = 12/9 = 4/3.

Now check the third term coefficient: $\frac{n(n-1)a^2}{2} = \frac{9(9-1)(4/3)^2}{2} = \frac{9 \times 8 \times (16/9)}{2} = \frac{72 \times (16/9)}{2} = \frac{8 \times 16}{2} = \frac{128}{2} = 64$. This matches the given third term coefficient.

So, the value of n is 9.

Revision Table: Binomial Expansion Concepts

Additional Information: Applications of Binomial Expansion

Binomial expansion is a fundamental concept in algebra with various applications:

• Probability: In probability, the binomial distribution uses binomial coefficients.
• Calculus: Used in Taylor series and Maclaurin series expansions of functions.
• Approximations: For small values of x, (1+x)ⁿ can be approximated using the first few terms (e.g., (1+x)ⁿ ≈ 1 + nx).
• Combinatorics: The coefficients themselves represent the number of ways to choose k items from a set of n items.

Understanding the structure of the terms in a binomial expansion, especially the first few terms, is crucial for solving problems like this one where specific terms are given.