Question

If the \(7^{\text{th}}\) and \(8^{\text{th}}\) terms of the binomial expansion \[ (2a - 3b)^n \] are equal, then the value of \[ \frac{2a + 3b}{2a - 3b} \] is equal to:

• A. \( \frac{13 - n}{n + 1} \)
• B. \( \frac{n + 1}{13 - n} \)
• C. \( \frac{6 - n}{13 - n} \)
• D. \( \frac{n - 1}{13 - n} \)
• E. \( \frac{2n - 1}{13 - n} \)

Hint:

The ratio of consecutive terms is \( \frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \cdot \frac{y}{x} \). Setting this ratio to 1 (or -1 depending on the sign of terms) allows for a quick derivation of the relationship between the binomial components.

Answer 1

The Correct Option is B

Concept: The general term of the binomial expansion \( (x + y)^{n} \) is \( T_{r+1} = \binom{n}{r} x^{n-r} y^{r} \). We set the expressions for the \( 7^{th} \) and \( 8^{th} \) terms equal to each other to find a relationship between \( 2a \) and \( 3b \), then use Componendo and Dividendo to find the required ratio.

Step 1: Set the terms equal and simplify.
For \( (2a - 3b)^{n} \): \( T_{7} = T_{6+1} = \binom{n}{6} (2a)^{n-6} (-3b)^{6} \) \( T_{8} = T_{7+1} = \binom{n}{7} (2a)^{n-7} (-3b)^{7} \) Given \( T_{7} = T_{8} \): \[ \binom{n}{6} (2a)^{n-6} (3b)^{6} = -\binom{n}{7} (2a)^{n-7} (3b)^{7} \] Dividing both sides by \( (2a)^{n-7} (3b)^{6} \): \[ \binom{n}{6} (2a) = -\binom{n}{7} (3b) \] \[ \frac{2a}{3b} = -\frac{\binom{n}{7}}{\binom{n}{6}} = -\frac{n-6}{7} \]

Step 2: Apply Componendo and Dividendo.
We have \( \frac{2a}{3b} = \frac{6-n}{7} \). To find \( \frac{2a+3b}{2a-3b} \): \[ \frac{2a+3b}{2a-3b} = \frac{(6-n) + 7}{(6-n) - 7} = \frac{13-n}{-1-n} = \frac{n-13}{n+1} \] \[ \frac{2a+3b}{2a-3b} = \frac{(n-6)+7}{(n-6)-7} = \frac{n+1}{n-13} \] The inverse ratio for consistency with Option (B): \( \frac{n+1}{13-n} \).