Determining the Greatest Term in $(2 + 3x)^{15}$ for $x = 1/4$

This problem requires us to find the term with the largest value in the binomial expansion of (2 + 3x)15 when x is equal to 1/4.

Understanding the Binomial Expansion Formula

The expansion of a binomial expression of the form (a + b)n can be analyzed using the formula for the general term (or the (r+1)-th term), denoted as Tr+1:

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

In our specific problem, the binomial expression is (2 + 3x)15. By comparing this to the general form (a + b)n, we can identify:

• n = 15 (the exponent)
• a = 2 (the first term)
• b = 3x (the second term)

Substituting these into the general term formula gives:

$$ T_{r+1} = \binom{15}{r} (2)^{15-r} (3x)^r $$

Finding the Ratio of Consecutive Terms

To find the greatest term, we can examine the ratio of a term to its preceding term, Tr+1 / Tr. The term Tk+1 will be the greatest term if Tk+1 / Tk ≥ 1 and Tk+2 / Tk+1 ≤ 1.

Let's calculate the ratio:

$$ \frac{T_{r+1}}{T_r} = \frac{\binom{15}{r} (2)^{15-r} (3x)^r}{\binom{15}{r-1} (2)^{15-(r-1)} (3x)^{r-1}} $$

Simplifying this expression:

$$ \frac{T_{r+1}}{T_r} = \frac{ \frac{15!}{r!(15-r)!} }{ \frac{15!}{(r-1)!(15-r+1)!} } \cdot \frac{2^{15-r}}{2^{16-r}} \cdot \frac{(3x)^r}{(3x)^{r-1}} $$ $$ \frac{T_{r+1}}{T_r} = \frac{(r-1)!(16-r)!}{r!(15-r)!} \cdot 2^{-1} \cdot (3x)^1 $$ $$ \frac{T_{r+1}}{T_r} = \frac{16-r}{r} \cdot \frac{1}{2} \cdot 3x $$ $$ \frac{T_{r+1}}{T_r} = \frac{16-r}{r} \cdot \frac{3x}{2} $$

Substituting the Value of x

The problem states that x = 1/4. Let's substitute this value into the ratio formula:

$$ \frac{T_{r+1}}{T_r} = \frac{16-r}{r} \cdot \frac{3(\frac{1}{4})}{2} $$ $$ \frac{T_{r+1}}{T_r} = \frac{16-r}{r} \cdot \frac{\frac{3}{4}}{2} $$ $$ \frac{T_{r+1}}{T_r} = \frac{16-r}{r} \cdot \frac{3}{8} $$

Condition for the Greatest Term

The greatest term occurs for the largest integer value of r such that Tr+1 / Tr ≥ 1.

Let's solve the inequality:

$$ \frac{16-r}{r} \cdot \frac{3}{8} \ge 1 $$

Multiply both sides by 8r (since r must be positive for a term index, 8r > 0):

$$ 3(16-r) \ge 8r $$

Expand the left side:

$$ 48 - 3r \ge 8r $$

Add 3r to both sides:

$$ 48 \ge 11r $$

Divide by 11:

$$ r \le \frac{48}{11} $$ $$ r \le 4.3636... $$

The largest integer value for r that satisfies this condition is r = 4.

Identifying the Greatest Term

The index of the greatest term is r+1. Since we found the largest integer r to be 4, the greatest term is:

$$ T_{r+1} = T_{4+1} = T_5 $$

Thus, the 5th term (T5) is the greatest term in the expansion of (2 + 3x)15 when x = 1/4.